NEB Class 11 · Past paper

Chemistry NEB Class 11 2073 question paper

The complete NEB Class 11 2073 exam paper for Chemistry, all 33 questions with solved model answers.

Past Papers2074 2073 2071 2070 Model

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120732 marksNumericalHideAnswer
Calculate the mass of i) 4 atoms of carbon ii) 3 molecules of hydrogen.
One mole ($6.022\times10^{23}$ particles) of carbon atoms weighs $12\ \text{g}$ and of $\ce{H2}$ weighs $2\ \text{g}$.

(i) 4 carbon atoms:

$$ \begin{aligned} \text{mass} &= \dfrac{4\times12}{6.022\times10^{23}} \ &= \mathbf{7.97\times10^{-23}\ g} \end{aligned} $$

(ii) 3 hydrogen molecules ($\ce{H2}$):

$$ \begin{aligned} \text{mass} &= \dfrac{3\times2}{6.022\times10^{23}} \ &= \mathbf{9.96\times10^{-24}\ g} \end{aligned} $$
220732 marksHideAnswer
State the law of constant composition giving a suitable example.
Law of constant (definite) composition: a pure chemical compound always contains the same elements combined in the same fixed proportion by mass, no matter how it is prepared or where it comes from.

Example: pure water always contains hydrogen and oxygen in the mass ratio $1:8$ (i.e. $2\ \text{g H} : 16\ \text{g O}$), whether obtained from a river, from rain, or made in the laboratory.
320732 marksHideAnswer
Define discontinuous solubility curve. What types of salts are responsible for such curves?
Discontinuous solubility curve: a solubility-versus-temperature curve that shows a sudden break (kink) at a particular temperature instead of rising smoothly. The break occurs where the solid salt changes from one hydrated (or crystalline) form to another.

Salts responsible: hydrated salts that exist in more than one crystalline form and change over at a definite temperature, e.g. Glauber's salt $\ce{Na2SO4.10H2O}$ (which becomes anhydrous $\ce{Na2SO4}$ above $32.4,^\circ\text{C}$) and $\ce{CaCl2.6H2O}$.
420732 marksHideAnswer
Mention any two important characters of each of the following: i) Efflorescent substance ii) Amorphous solid.
(i) Efflorescent substance: a hydrated salt that loses its water of crystallisation to the air on standing. (a) It becomes powdery/opaque on the surface, and (b) its mass decreases on exposure. Example: $\ce{Na2CO3.10H2O}$.

(ii) Amorphous solid: (a) it has no regular long-range arrangement of particles (only short-range order), and (b) it has no sharp melting point (softens over a range) and is isotropic. Example: glass.
520732 marksHideAnswer
Assign the values of the quantum numbers n, l and m for the outermost electron in the sodium atom.
Sodium ($Z=11$): $\ce{1s^2 2s^2 2p^6 3s^1}$. The outermost electron is $3s^1$. - Principal $n=3$ - Azimuthal $l=0$ (for an $s$ orbital) - Magnetic $m=0$
620732 marksHideAnswer
Write down the electronic configuration of i) Al^3+ ii) S.
(i) $\ce{Al^{3+}}$: aluminium ($Z=13$) loses 3 electrons: $\ce{1s^2 2s^2 2p^6}$ (10 electrons, neon configuration). (ii) S ($Z=16$): $\ce{1s^2 2s^2 2p^6 3s^2 3p^4}$.
720732 marksHideAnswer
Define nuclear fusion reaction giving an example.
Nuclear fusion: the process in which two light nuclei combine to form a heavier nucleus, releasing a very large amount of energy (the mass lost appears as energy, $E=mc^2$).

Example: $$\ce{^2_1H + ^3_1H -> ^4_2He + ^1_0n} + \text{energy}$$ This is the source of the sun's energy.
820732 marksHideAnswer
Draw the Lewis structure of i) H2SO3 ii) N2O.
(i) $\ce{H2SO3}$ (sulphurous acid): two $\ce{-OH}$ groups and one doubly bonded O are attached to the central S, which also has one lone pair. Skeleton $\ce{H-O-S(=O)-O-H}$ with a lone pair on S. (ii) $\ce{N2O}$ (nitrous oxide): a linear...
920732 marksHideAnswer
Each carbon-oxygen bond in CO2 is polar but the CO2 molecule is non-polar. Explain the proper reason.
Each $\ce{C=O}$ bond is polar because oxygen is more electronegative than carbon, so every bond has a dipole moment directed towards O.

However, $\ce{CO2}$ is a linear molecule ($\ce{O=C=O}$, bond angle $180^\circ$). The two bond dipoles are equal in magnitude but exactly opposite in direction, so they cancel each other. The net (resultant) dipole moment is zero, and hence the molecule as a whole is non-polar.
1020732 marksHideAnswer
Predict which of the following pair has larger electron affinity and why: O and F.
Fluorine (F) has the larger electron affinity than oxygen. Across a period, effective nuclear charge increases and atomic size decreases, so the incoming electron is held more strongly. Fluorine, being further to the right with a higher ...
1120732 marksNumericalHideAnswer
Define oxidation number. What is the oxidation number of P in NaH2PO2?
Oxidation number: the apparent charge that an atom would have if all the bonding electron pairs were assigned to the more electronegative atom of each bond.

P in $\ce{NaH2PO2}$ (sodium hypophosphite):

Na $=+1$, each H $=+1$, each O $=-2$.

$$ \begin{aligned} (+1) + 2(+1) + x + 2(-2) &= 0 \ 1 + 2 + x - 4 &= 0 \ x &= +1 \end{aligned} $$

So P $=+1$.
1220732 marksHideAnswer
Write the expression for the equilibrium constant (Kc) for 4NO(g) + 6H2O(g) <=> 4NH3(g) + 5O2(g). What is the relationship between Kp and Kc for the above reaction?
$$ \begin{aligned} Kc &= \dfrac{[\ce{NH3}]^4,[\ce{O2}]^5}{[\ce{NO}]^4,[\ce{H2O}]^6} \ \Delta n &= (4+5)-(4+6) \ &= 9-10 \ &= -1 \end{aligned} $$ Therefore $$ \begin{aligned} Kp &= Kc(RT)^{\Delta n} \ &= Kc(RT)^{-1} \ &= \dfrac{Kc}...
1320732 marksHideAnswer
What is nascent hydrogen? Mention an example to show that nascent hydrogen is a more powerful reducing agent than molecular hydrogen.
Nascent hydrogen: hydrogen in its freshly generated atomic form ($\ce{[H]}$, "at the moment of birth"), produced in the reaction mixture itself (e.g. from Zn + acid). It is far more reactive than ordinary molecular hydrogen $\ce{H2}$.

Example: nascent hydrogen (from $\ce{Zn + H2SO4}$) reduces acidified $\ce{KMnO4}$ (decolourises it) and reduces ferric to ferrous ions: $$\ce{2FeCl3 + 2[H] -> 2FeCl2 + 2HCl}$$ Molecular $\ce{H2}$ does not carry out these reductions under the same conditions, showing nascent hydrogen is the stronger reducing agent.
1420732 marksHideAnswer
How does the formation of ozone take place in the stratosphere? Give any two uses of ozone.
In the stratosphere, high-energy ultraviolet radiation from the sun first splits oxygen molecules into free atoms,

$$\ce{O2 ->[UV] 2O}$$

and each of these atoms then combines with another oxygen molecule to form ozone,

$$\ce{O + O2 -> O3}$$

This ozone layer absorbs harmful UV rays and protects life on earth.

Ozone has two important uses: it sterilises and purifies drinking water and air, and it acts as an oxidising and bleaching agent for oils, ivory and starch.
1520732 marksHideAnswer
What happens when the gas obtained by heating methanoic acid with conc. sulphuric acid is heated with Fe2O3?
Heating methanoic (formic) acid with conc. $\ce{H2SO4}$ gives carbon monoxide: $$\ce{HCOOH ->[conc.\ H2SO4] CO(g) + H2O}$$ This $\ce{CO}$, being a good reducing agent, reduces heated ferric oxide to iron: $$\ce{Fe2O3 + 3CO ->[\Delta] 2Fe + 3CO2(g)}$$
1620732 marksHideAnswer
How is SO2 responsible for acid rain?
The $\ce{SO2}$ released from burning coal and fossil fuels is oxidised in the atmosphere to $\ce{SO3}$, $$\ce{2SO2 + O2 - 2SO3}$$ which then dissolves in rain water to form sulphuric acid, $$\ce{SO3 + H2O - H2SO4}$$ The $\ce{SO2}$ itself...
1720732 marksHideAnswer
'Every ore is a mineral but every mineral is not an ore.' Give reason.
A mineral is any naturally occurring substance containing a metal in combined (or free) form. An ore is a mineral from which the metal can be extracted conveniently and profitably. So all ores are minerals, but a mineral becomes an ore o...
1820732 marksHideAnswer
Give a balanced chemical reaction for the preparation of slaked lime. How would you obtain lime water from slaked lime?
Preparation of slaked lime (adding water to quicklime, a vigorous exothermic reaction): $$\ce{CaO + H2O - Ca(OH)2} + \text{heat}$$ Lime water: a small amount of slaked lime, $\ce{Ca(OH)2}$, is shaken with excess water and the mixture is ...
1920732 marksHideAnswer
Give the structure of the following compounds: i) 2,3-dimethylbut-2-ene ii) ethyl ethanoate.
(i) 2,3-dimethylbut-2-ene: a four-carbon chain with a double bond between C-2 and C-3 and a methyl group on each of C-2 and C-3: $$\ce{(CH3)2C=C(CH3)2}$$ (ii) Ethyl ethanoate (ethyl acetate): the ester of ethanoic acid and ethanol: $$\ce...
2020732 marksHideAnswer
How is an electrophile different from a nucleophile? Give a suitable example of each.
Electrophile: an electron-deficient (electron-loving) species that seeks and accepts an electron pair; it is positively charged or has an incomplete octet. Example: $\ce{NO2+}$, $\ce{H+}$, $\ce{BF3}$. Nucleophile: an electron-rich (nucle...
2120732 marksHideAnswer
Identify the major products A and B in the following reaction sequence and give their IUPAC names: CH3CH2Br --(alc. KOH, heat)--> A --(ozonolysis)--> B.
Alcoholic KOH first removes HBr from the bromoethane (dehydrohalogenation), leaving a double bond. $$\ce{CH3CH2Br -[alc.\ KOH][\Delta] CH2=CH2 + HBr}$$ So A is ethene. Ozonolysis then cleaves that double bond into carbonyl compounds, and...
2220732 marksHideAnswer
How would you prepare methane from i) chloromethane ii) sodium ethanoate?
(i) From chloromethane (reduction by nascent hydrogen, Zn/HCl couple): $$\ce{CH3Cl + 2[H] -[Zn/Cu] CH4 + HCl}$$ (ii) From sodium ethanoate (decarboxylation by heating with soda lime, NaOH + CaO): $$\ce{CH3COONa + NaOH -[CaO][\Delta] CH4 ...
2320735 marksHideAnswer
What property did Mendeleev use to classify the elements in his periodic table? Point out the anomalies in Mendeleev's periodic table.
Basis: Mendeleev classified the elements in order of their increasing atomic mass, grouping together elements with similar properties (his law: properties are a periodic function of atomic mass).

Anomalies (defects):
1. Position of hydrogen was not fixed (it resembles both Group I and Group VII).
2. Anomalous pairs: in some places a heavier element was placed before a lighter one to keep similar elements together, e.g. Ar (40) before K (39), Co before Ni, Te before I.
3. Isotopes had no separate place (different masses, same element).
4. Grouping of dissimilar elements together in sub-groups (e.g. coinage metals with alkali metals).
5. The position of the lanthanides and actinides could not be accommodated.
2420735 marksHideAnswer
Balance the following equation by the oxidation number or ion-electron method and point out the oxidant and reductant: Zn + NaNO3 + NaOH -> Na2ZnO2 + NH3 + H2O.
Zn goes $0\to+2$ (loses $2e^-$); N goes $+5\to-3$ (gains $8e^-$). To balance electrons, 4 Zn ($8e^-$ lost) per 1 N ($8e^-$ gained): $$\ce{4Zn + NaNO3 + 7NaOH - 4Na2ZnO2 + NH3 + 2H2O}$$ Check: Na $$ \begin{aligned} 1+7 &= 8 \ &= 4\times2...
2520735 marksHideAnswer
How is ethene prepared in the laboratory? What happens when ethene is passed through Baeyer's reagent?
Laboratory preparation of ethene (dehydration of ethanol by heating with excess conc. $\ce{H2SO4}$ at ~$170,^\circ\text{C}$): $$\ce{C2H5OH -[conc.\ H2SO4][170,^\circ C] C2H4(g) + H2O}$$ The gas is washed and collected over water. With ...
2620735 marksHideAnswer
Starting from sulphur, how would you obtain sulphuric acid (reactions only)? How does conc. H2SO4 react with i) sugar ii) P4? Why is conc. sulphuric acid diluted by adding acid to water and not water to acid?
Sulphur to sulphuric acid (contact process, reactions only): $$ \begin{aligned} \ce{S + O2 - SO2} \ \ce{2SO2 + O2 <=[V2O5] 2SO3} \ \ce{SO3 + H2SO4 - H2S2O7} \ \ce{H2S2O7 + H2O - 2H2SO4} \end{aligned} $$ i) With sugar (dehydration to b...
2720735 marksHideAnswer
Differentiate between i) calcination and roasting ii) carbon-reduction process and thermite process.
i) Calcination vs roasting:
- Calcination is heating an ore strongly in limited/absence of air to expel moisture and volatile matter and to convert carbonates/hydroxides to oxides, e.g.
$$ \ce{CaCO3 ->[\Delta] CaO + CO2} $$

.
- Roasting is heating an ore strongly in a current of air, mainly for sulphide ores, converting them to oxides with release of $\ce{SO2}$, e.g.
$$ \ce{2ZnS + 3O2 -> 2ZnO + 2SO2} $$

.

ii) Carbon-reduction vs thermite process:
- Carbon reduction uses coke to reduce a metal oxide, e.g.
$$ \ce{ZnO + C -> Zn + CO} $$

; suitable for moderately reactive metals.
- Thermite (alumino-thermite) process uses aluminium as the reducing agent for oxides not easily reduced by carbon, e.g.
$$ \ce{Fe2O3 + 2Al -> 2Fe + Al2O3} $$

; highly exothermic, used to weld rails.
2820735 marksHideAnswer
Describe the preparation of bromine from carnallite. What happens when bromine is passed through hot and conc. NaOH?
Preparation from carnallite: the bromide-rich bittern (mother liquor from carnallite $\ce{KCl.MgCl2.6H2O}$) is acidified and warmed, and chlorine is passed through it; chlorine displaces bromine, which is swept out by air/steam and condensed: $$\ce{MgBr2 + Cl2 -> MgCl2 + Br2(g)}$$

Bromine with hot conc. NaOH (disproportionation to bromide and bromate): $$\ce{3Br2 + 6NaOH(hot,\ conc) -> 5NaBr + NaBrO3 + 3H2O}$$ (With cold dilute NaOH it gives $\ce{NaBr}$ + $\ce{NaBrO}$ instead.)
2920735 marksNumericalHideAnswer
For the reaction 2NaHCO3 --(heat)--> Na2CO3 + H2O + CO2, 2.5 g of a sample of NaHCO3 when strongly heated gives 310 cc of CO2 at 27 degree C and 760 mmHg. i) Calculate the percentage purity of the sample. ii) How many moles of water are produced? iii) What mass of pure HCl is required to neutralize the Na2CO3 produced?
$$ \begin{aligned} \ce{2NaHCO3 ->[\Delta] Na2CO3 + H2O + CO2} \ n_{\ce{CO2}} &= \dfrac{PV}{RT} \ &= \dfrac{1\times0.310}{0.0821\times300} \ &= 0.01259\ \text{mol} \end{aligned} $$

i) Purity:

$$ \ce{2NaHCO3 -> 1CO2} $$

, so

$$ \begin{aligned} \text{pure }\ce{NaHCO3} &= 2\times0.01259 \ &= 0.02517\ \text{mol} \ &= 0.02517\times84 \ &= 2.115\ \text{g} \ %\ \text{purity} &= \dfrac{2.115}{2.5}\times100 \ &= \mathbf{84.6%} \end{aligned} $$

ii) Water:

$1\ \ce{H2O}$ per $1\ \ce{CO2}$, so $n_{\ce{H2O}}=\mathbf{0.0126\ mol}$.

iii) HCl to neutralise $\ce{Na2CO3}$:

$n_{\ce{Na2CO3}}=0.01259$ mol.

$$ \begin{aligned} \ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2} \ n_{\ce{HCl}} &= 2\times0.01259 \ &= 0.02517\ \text{mol} \ \text{mass} &= 0.02517\times36.5 \ &= \mathbf{0.919\ g} \end{aligned} $$
30207310 marksNumericalHideAnswer
State Charle's law. How did this law lead to the development of the absolute temperature scale? Plot the temperature-volume relationship indicating absolute zero. A saturated hydrocarbon (CnH2n+2) diffuses through a porous membrane twice as fast as sulphur dioxide. Determine the molecular formula of the hydrocarbon.
Charles's law: at constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature: $V\propto T$, i.e. $\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$.

Absolute scale: experimentally the volume falls by $\tfrac{1}{273}$ of its $0,^\circ\text{C}$ value per degree of cooling. Extending the $V$-$t$ line, $V$ would become zero at $-273,^\circ\text{C}$; a lower temperature (negative volume) is impossible, so this is absolute zero and the Kelvin scale starts there: $T(\text{K})=t(^\circ\text{C})+273$.

Plot: a straight line of $V$ (y-axis) against $t\ (^\circ\text{C})$ (x-axis) that, when extrapolated backwards, meets the temperature axis at $-273,^\circ\text{C}$ (where $V=0$), which marks absolute zero.

Molecular formula (Graham's law):

$$ \begin{aligned} \dfrac{r_{HC}}{r_{\ce{SO2}}} &= 2 \ &= \sqrt{\dfrac{M_{\ce{SO2}}}{M_{HC}}} \ 4 &= \dfrac{64}{M_{HC}} \ M_{HC} &= 16 \ 14n+2 &= 16 \ &\Rightarrow n = 1 \end{aligned} $$

The hydrocarbon is methane, $\ce{CH4}$.
31207310 marksHideAnswer
Sketch a well-labelled diagram for the manufacture of caustic soda (NaOH) by Castner-Kellner's process. Explain the principle and procedure involved. What happens when caustic soda is i) heated with sulphur ii) treated with zinc? How would you convert caustic soda into sodium silicate?
Castner-Kellner process: brine is electrolysed with a flowing mercury cathode and graphite anodes. $$ \begin{aligned} \text{Anode: }\ce{2Cl- - Cl2(g) + 2e-} \ \text{Cathode: }\ce{Na+ + e- - Na}\ (\to \text{sodium amalgam}) \end{aligned}...
32207310 marksHideAnswer
Write the principle and process involved along with a self-explanatory diagram for the manufacture of nitric acid by catalytic oxidation of ammonia. How does concentrated nitric acid react with i) iron ii) iodine? Why is conc. nitric acid stored in a dark bottle?
Ostwald's process (catalytic oxidation of ammonia) - principle and process: $$ \begin{aligned} \ce{4NH3(g) + 5O2(g) ->[Pt/Rh][~800,^\circ C] 4NO(g) + 6H2O(g)} \ \ce{2NO(g) + O2(g) -> 2NO2(g)} \ \ce{4NO2(g) + O2(g) + 2H2O(l) -> 4HNO3(aq)} \end{aligned} $$ A $1:8$ ammonia-air mixture is passed over Pt/Rh gauze (~$800,^\circ\text{C}$); the $\ce{NO}$ is cooled, oxidised to $\ce{NO2}$ by more air, and absorbed in water in a tower to give nitric acid.

i) Conc. $\ce{HNO3}$ + iron: cold conc. $\ce{HNO3}$ makes iron passive, forming a thin protective oxide layer that stops further reaction (so it is transported in iron/steel tankers).

ii) Conc. $\ce{HNO3}$ + iodine (oxidises $\ce{I2}$ to iodic acid): $$\ce{I2 + 10HNO3 -> 2HIO3 + 10NO2(g) + 4H2O}$$

Dark bottle: nitric acid is decomposed by light (and heat) into brown $\ce{NO2}$, which colours it yellow: $$\ce{4HNO3 ->[light] 4NO2 + O2 + 2H2O}$$ A dark (amber) bottle keeps out light and prevents this decomposition.
33207310 marksHideAnswer
Write short notes on any two: i) Rutherford atomic model ii) Le-Chatelier's principle and its applications iii) Characteristics of homologous series iv) Lassaigne's test for nitrogen.
i) Rutherford atomic model: from the $\alpha$-scattering experiment, Rutherford proposed that an atom has a tiny, dense, positively charged nucleus carrying almost all its mass, with electrons revolving around it in the mostly empty space; the atom is electrically neutral. Its defect: it could not explain atomic stability or line spectra.

ii) Le-Chatelier's principle: if a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to undo the change. Applications: in Haber's process ($\ce{N2 + 3H2 <=> 2NH3}$) high pressure and low temperature increase $\ce{NH3}$ yield; in the contact process high pressure/excess $\ce{O2}$ increase $\ce{SO3}$.

iii) Characteristics of homologous series: same general formula and same functional group; successive members differ by $\ce{-CH2-}$ ($14\ \text{u}$); gradual gradation of physical properties with molar mass; similar chemical properties; all made by similar general methods.

iv) Lassaigne's test for nitrogen: the compound is fused with sodium to give $\ce{NaCN}$; the extract is boiled with $\ce{FeSO4}$ and acidified with $\ce{HCl}$; a Prussian-blue colour confirms nitrogen:

$$ \ce{6NaCN + FeSO4 -> Na4[Fe(CN)6] + Na2SO4} $$

;

$$ \ce{3Na4[Fe(CN)6] + 4FeCl3 -> Fe4[Fe(CN)6]3 + 12NaCl} $$

.

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NEB Class 11 · Past paper

Chemistry NEB Class 11 2073 question paper

The complete NEB Class 11 2073 exam paper for Chemistry, all 33 questions with solved model answers.

Past Papers2074 2073 2071 2070 Model

Tap a question to open its answer.

120732 marksNumericalHideAnswer
Calculate the mass of i) 4 atoms of carbon ii) 3 molecules of hydrogen.
One mole ( $6.022\times10^{23}$ particles) of carbon atoms weighs $12\ \text{g}$ and of $\ce{H2}$ weighs $2\ \text{g}$ .

(i) 4 carbon atoms:
$\begin{aligned} \text{mass} &= \dfrac{4\times12}{6.022\times10^{23}} \\ &= \mathbf{7.97\times10^{-23}\ g} \end{aligned}$
(ii) 3 hydrogen molecules ( $\ce{H2}$ ):
$\begin{aligned} \text{mass} &= \dfrac{3\times2}{6.022\times10^{23}} \\ &= \mathbf{9.96\times10^{-24}\ g} \end{aligned}$
One mole ($6.022\times10^{23}$ particles) of carbon atoms weighs $12\ \text{g}$ and of $\ce{H2}$ weighs $2\ \text{g}$.

(i) 4 carbon atoms:

$$ \begin{aligned} \text{mass} &= \dfrac{4\times12}{6.022\times10^{23}} \ &= \mathbf{7.97\times10^{-23}\ g} \end{aligned} $$

(ii) 3 hydrogen molecules ($\ce{H2}$):

$$ \begin{aligned} \text{mass} &= \dfrac{3\times2}{6.022\times10^{23}} \ &= \mathbf{9.96\times10^{-24}\ g} \end{aligned} $$
220732 marksHideAnswer
State the law of constant composition giving a suitable example.
Law of constant (definite) composition: a pure chemical compound always contains the same elements combined in the same fixed proportion by mass, no matter how it is prepared or where it comes from.

Example: pure water always contains hydrogen and oxygen in the mass ratio $1:8$ (i.e. $2\ \text{g H} : 16\ \text{g O}$ ), whether obtained from a river, from rain, or made in the laboratory.
Law of constant (definite) composition: a pure chemical compound always contains the same elements combined in the same fixed proportion by mass, no matter how it is prepared or where it comes from.

Example: pure water always contains hydrogen and oxygen in the mass ratio $1:8$ (i.e. $2\ \text{g H} : 16\ \text{g O}$), whether obtained from a river, from rain, or made in the laboratory.
320732 marksHideAnswer
Define discontinuous solubility curve. What types of salts are responsible for such curves?
Discontinuous solubility curve: a solubility-versus-temperature curve that shows a sudden break (kink) at a particular temperature instead of rising smoothly. The break occurs where the solid salt changes from one hydrated (or crystalline) form to another.

Salts responsible: hydrated salts that exist in more than one crystalline form and change over at a definite temperature, e.g. Glauber's salt $\ce{Na2SO4.10H2O}$ (which becomes anhydrous $\ce{Na2SO4}$ above $32.4\,^\circ\text{C}$ ) and $\ce{CaCl2.6H2O}$ .
Discontinuous solubility curve: a solubility-versus-temperature curve that shows a sudden break (kink) at a particular temperature instead of rising smoothly. The break occurs where the solid salt changes from one hydrated (or crystalline) form to another.

Salts responsible: hydrated salts that exist in more than one crystalline form and change over at a definite temperature, e.g. Glauber's salt $\ce{Na2SO4.10H2O}$ (which becomes anhydrous $\ce{Na2SO4}$ above $32.4,^\circ\text{C}$) and $\ce{CaCl2.6H2O}$.
420732 marksHideAnswer
Mention any two important characters of each of the following: i) Efflorescent substance ii) Amorphous solid.
(i) Efflorescent substance: a hydrated salt that loses its water of crystallisation to the air on standing. (a) It becomes powdery/opaque on the surface, and (b) its mass decreases on exposure. Example: $\ce{Na2CO3.10H2O}$ .

(ii) Amorphous solid: (a) it has no regular long-range arrangement of particles (only short-range order), and (b) it has no sharp melting point (softens over a range) and is isotropic. Example: glass.
(i) Efflorescent substance: a hydrated salt that loses its water of crystallisation to the air on standing. (a) It becomes powdery/opaque on the surface, and (b) its mass decreases on exposure. Example: $\ce{Na2CO3.10H2O}$.

(ii) Amorphous solid: (a) it has no regular long-range arrangement of particles (only short-range order), and (b) it has no sharp melting point (softens over a range) and is isotropic. Example: glass.
520732 marksHideAnswer
Assign the values of the quantum numbers n, l and m for the outermost electron in the sodium atom.
Sodium ( $Z=11$ ): $\ce{1s^2 2s^2 2p^6 3s^1}$ . The outermost electron is $3s^1$ . - Principal $n=3$ - Azimuthal $l=0$ (for an $s$ orbital) - Magnetic $m=0$
Sodium ($Z=11$): $\ce{1s^2 2s^2 2p^6 3s^1}$. The outermost electron is $3s^1$. - Principal $n=3$ - Azimuthal $l=0$ (for an $s$ orbital) - Magnetic $m=0$
620732 marksHideAnswer
Write down the electronic configuration of i) Al^3+ ii) S.
(i) $\ce{Al^{3+}}$ : aluminium ( $Z=13$ ) loses 3 electrons: $\ce{1s^2 2s^2 2p^6}$ (10 electrons, neon configuration). (ii) S ( $Z=16$ ): $\ce{1s^2 2s^2 2p^6 3s^2 3p^4}$ .
(i) $\ce{Al^{3+}}$: aluminium ($Z=13$) loses 3 electrons: $\ce{1s^2 2s^2 2p^6}$ (10 electrons, neon configuration). (ii) S ($Z=16$): $\ce{1s^2 2s^2 2p^6 3s^2 3p^4}$.
720732 marksHideAnswer
Define nuclear fusion reaction giving an example.
Nuclear fusion: the process in which two light nuclei combine to form a heavier nucleus, releasing a very large amount of energy (the mass lost appears as energy, $E=mc^2$ ).

Example: $\ce{^2_1H + ^3_1H -> ^4_2He + ^1_0n} + \text{energy}$ This is the source of the sun's energy.
Nuclear fusion: the process in which two light nuclei combine to form a heavier nucleus, releasing a very large amount of energy (the mass lost appears as energy, $E=mc^2$).

Example: $$\ce{^2_1H + ^3_1H -> ^4_2He + ^1_0n} + \text{energy}$$ This is the source of the sun's energy.
820732 marksHideAnswer
Draw the Lewis structure of i) H2SO3 ii) N2O.
(i) $\ce{H2SO3}$ (sulphurous acid): two $\ce{-OH}$ groups and one doubly bonded O are attached to the central S, which also has one lone pair. Skeleton $\ce{H-O-S(=O)-O-H}$ with a lone pair on S. (ii) $\ce{N2O}$ (nitrous oxide): a linear...
(i) $\ce{H2SO3}$ (sulphurous acid): two $\ce{-OH}$ groups and one doubly bonded O are attached to the central S, which also has one lone pair. Skeleton $\ce{H-O-S(=O)-O-H}$ with a lone pair on S. (ii) $\ce{N2O}$ (nitrous oxide): a linear...
920732 marksHideAnswer
Each carbon-oxygen bond in CO2 is polar but the CO2 molecule is non-polar. Explain the proper reason.
Each $\ce{C=O}$ bond is polar because oxygen is more electronegative than carbon, so every bond has a dipole moment directed towards O.

However, $\ce{CO2}$ is a linear molecule ( $\ce{O=C=O}$ , bond angle $180^\circ$ ). The two bond dipoles are equal in magnitude but exactly opposite in direction, so they cancel each other. The net (resultant) dipole moment is zero, and hence the molecule as a whole is non-polar.

Each $\ce{C=O}$ bond is polar because oxygen is more electronegative than carbon, so every bond has a dipole moment directed towards O.

However, $\ce{CO2}$ is a linear molecule ($\ce{O=C=O}$, bond angle $180^\circ$). The two bond dipoles are equal in magnitude but exactly opposite in direction, so they cancel each other. The net (resultant) dipole moment is zero, and hence the molecule as a whole is non-polar.
1020732 marksHideAnswer
Predict which of the following pair has larger electron affinity and why: O and F.
Fluorine (F) has the larger electron affinity than oxygen. Across a period, effective nuclear charge increases and atomic size decreases, so the incoming electron is held more strongly. Fluorine, being further to the right with a higher ...
1120732 marksNumericalHideAnswer
Define oxidation number. What is the oxidation number of P in NaH2PO2?
Oxidation number: the apparent charge that an atom would have if all the bonding electron pairs were assigned to the more electronegative atom of each bond.

P in $\ce{NaH2PO2}$ (sodium hypophosphite):

Na $=+1$ , each H $=+1$ , each O $=-2$ .
$\begin{aligned} (+1) + 2(+1) + x + 2(-2) &= 0 \\ 1 + 2 + x - 4 &= 0 \\ x &= +1 \end{aligned}$
So P $=+1$ .
Oxidation number: the apparent charge that an atom would have if all the bonding electron pairs were assigned to the more electronegative atom of each bond.

P in $\ce{NaH2PO2}$ (sodium hypophosphite):

Na $=+1$, each H $=+1$, each O $=-2$.

$$ \begin{aligned} (+1) + 2(+1) + x + 2(-2) &= 0 \ 1 + 2 + x - 4 &= 0 \ x &= +1 \end{aligned} $$

So P $=+1$.
1220732 marksHideAnswer
Write the expression for the equilibrium constant (Kc) for 4NO(g) + 6H2O(g) <=> 4NH3(g) + 5O2(g). What is the relationship between Kp and Kc for the above reaction?
$\begin{aligned} Kc &= \dfrac{[\ce{NH3}]^4\,[\ce{O2}]^5}{[\ce{NO}]^4\,[\ce{H2O}]^6} \\ \Delta n &= (4+5)-(4+6) \\ &= 9-10 \\ &= -1 \end{aligned}$ Therefore $$ \begin{aligned} Kp &= Kc(RT)^{\Delta n} \ &= Kc(RT)^{-1} \ &= \dfrac{Kc}...
$$ \begin{aligned} Kc &= \dfrac{[\ce{NH3}]^4,[\ce{O2}]^5}{[\ce{NO}]^4,[\ce{H2O}]^6} \ \Delta n &= (4+5)-(4+6) \ &= 9-10 \ &= -1 \end{aligned} $$ Therefore $$ \begin{aligned} Kp &= Kc(RT)^{\Delta n} \ &= Kc(RT)^{-1} \ &= \dfrac{Kc}...
1320732 marksHideAnswer
What is nascent hydrogen? Mention an example to show that nascent hydrogen is a more powerful reducing agent than molecular hydrogen.
Nascent hydrogen: hydrogen in its freshly generated atomic form ( $\ce{[H]}$ , "at the moment of birth"), produced in the reaction mixture itself (e.g. from Zn + acid). It is far more reactive than ordinary molecular hydrogen $\ce{H2}$ .

Example: nascent hydrogen (from $\ce{Zn + H2SO4}$ ) reduces acidified $\ce{KMnO4}$ (decolourises it) and reduces ferric to ferrous ions: $\ce{2FeCl3 + 2[H] -> 2FeCl2 + 2HCl}$ Molecular $\ce{H2}$ does not carry out these reductions under the same conditions, showing nascent hydrogen is the stronger reducing agent.
Nascent hydrogen: hydrogen in its freshly generated atomic form ($\ce{[H]}$, "at the moment of birth"), produced in the reaction mixture itself (e.g. from Zn + acid). It is far more reactive than ordinary molecular hydrogen $\ce{H2}$.

Example: nascent hydrogen (from $\ce{Zn + H2SO4}$) reduces acidified $\ce{KMnO4}$ (decolourises it) and reduces ferric to ferrous ions: $$\ce{2FeCl3 + 2[H] -> 2FeCl2 + 2HCl}$$ Molecular $\ce{H2}$ does not carry out these reductions under the same conditions, showing nascent hydrogen is the stronger reducing agent.
1420732 marksHideAnswer
How does the formation of ozone take place in the stratosphere? Give any two uses of ozone.
In the stratosphere, high-energy ultraviolet radiation from the sun first splits oxygen molecules into free atoms,

$\ce{O2 ->[UV] 2O}$

and each of these atoms then combines with another oxygen molecule to form ozone,

$\ce{O + O2 -> O3}$

This ozone layer absorbs harmful UV rays and protects life on earth.

Ozone has two important uses: it sterilises and purifies drinking water and air, and it acts as an oxidising and bleaching agent for oils, ivory and starch.
In the stratosphere, high-energy ultraviolet radiation from the sun first splits oxygen molecules into free atoms,

$$\ce{O2 ->[UV] 2O}$$

and each of these atoms then combines with another oxygen molecule to form ozone,

$$\ce{O + O2 -> O3}$$

This ozone layer absorbs harmful UV rays and protects life on earth.

Ozone has two important uses: it sterilises and purifies drinking water and air, and it acts as an oxidising and bleaching agent for oils, ivory and starch.
1520732 marksHideAnswer
What happens when the gas obtained by heating methanoic acid with conc. sulphuric acid is heated with Fe2O3?
Heating methanoic (formic) acid with conc. $\ce{H2SO4}$ gives carbon monoxide: $\ce{HCOOH ->[conc.\ H2SO4] CO(g) + H2O}$ This $\ce{CO}$ , being a good reducing agent, reduces heated ferric oxide to iron: $\ce{Fe2O3 + 3CO ->[\Delta] 2Fe + 3CO2(g)}$
Heating methanoic (formic) acid with conc. $\ce{H2SO4}$ gives carbon monoxide: $$\ce{HCOOH ->[conc.\ H2SO4] CO(g) + H2O}$$ This $\ce{CO}$, being a good reducing agent, reduces heated ferric oxide to iron: $$\ce{Fe2O3 + 3CO ->[\Delta] 2Fe + 3CO2(g)}$$
1620732 marksHideAnswer
How is SO2 responsible for acid rain?
The $\ce{SO2}$ released from burning coal and fossil fuels is oxidised in the atmosphere to $\ce{SO3}$ , $\ce{2SO2 + O2 - 2SO3}$ which then dissolves in rain water to form sulphuric acid, $\ce{SO3 + H2O - H2SO4}$ The $\ce{SO2}$ itself...
The $\ce{SO2}$ released from burning coal and fossil fuels is oxidised in the atmosphere to $\ce{SO3}$, $$\ce{2SO2 + O2 - 2SO3}$$ which then dissolves in rain water to form sulphuric acid, $$\ce{SO3 + H2O - H2SO4}$$ The $\ce{SO2}$ itself...
1720732 marksHideAnswer
'Every ore is a mineral but every mineral is not an ore.' Give reason.
A mineral is any naturally occurring substance containing a metal in combined (or free) form. An ore is a mineral from which the metal can be extracted conveniently and profitably. So all ores are minerals, but a mineral becomes an ore o...
1820732 marksHideAnswer
Give a balanced chemical reaction for the preparation of slaked lime. How would you obtain lime water from slaked lime?
Preparation of slaked lime (adding water to quicklime, a vigorous exothermic reaction): $\ce{CaO + H2O - Ca(OH)2} + \text{heat}$ Lime water: a small amount of slaked lime, $\ce{Ca(OH)2}$ , is shaken with excess water and the mixture is ...
Preparation of slaked lime (adding water to quicklime, a vigorous exothermic reaction): $$\ce{CaO + H2O - Ca(OH)2} + \text{heat}$$ Lime water: a small amount of slaked lime, $\ce{Ca(OH)2}$, is shaken with excess water and the mixture is ...
1920732 marksHideAnswer
Give the structure of the following compounds: i) 2,3-dimethylbut-2-ene ii) ethyl ethanoate.
(i) 2,3-dimethylbut-2-ene: a four-carbon chain with a double bond between C-2 and C-3 and a methyl group on each of C-2 and C-3: $\ce{(CH3)2C=C(CH3)2}$ (ii) Ethyl ethanoate (ethyl acetate): the ester of ethanoic acid and ethanol: $$\ce...
(i) 2,3-dimethylbut-2-ene: a four-carbon chain with a double bond between C-2 and C-3 and a methyl group on each of C-2 and C-3: $$\ce{(CH3)2C=C(CH3)2}$$ (ii) Ethyl ethanoate (ethyl acetate): the ester of ethanoic acid and ethanol: $$\ce...
2020732 marksHideAnswer
How is an electrophile different from a nucleophile? Give a suitable example of each.
Electrophile: an electron-deficient (electron-loving) species that seeks and accepts an electron pair; it is positively charged or has an incomplete octet. Example: $\ce{NO2+}$ , $\ce{H+}$ , $\ce{BF3}$ . Nucleophile: an electron-rich (nucle...
Electrophile: an electron-deficient (electron-loving) species that seeks and accepts an electron pair; it is positively charged or has an incomplete octet. Example: $\ce{NO2+}$, $\ce{H+}$, $\ce{BF3}$. Nucleophile: an electron-rich (nucle...
2120732 marksHideAnswer
Identify the major products A and B in the following reaction sequence and give their IUPAC names: CH3CH2Br --(alc. KOH, heat)--> A --(ozonolysis)--> B.
Alcoholic KOH first removes HBr from the bromoethane (dehydrohalogenation), leaving a double bond. $\ce{CH3CH2Br -[alc.\ KOH][\Delta] CH2=CH2 + HBr}$ So A is ethene. Ozonolysis then cleaves that double bond into carbonyl compounds, and...
Alcoholic KOH first removes HBr from the bromoethane (dehydrohalogenation), leaving a double bond. $$\ce{CH3CH2Br -[alc.\ KOH][\Delta] CH2=CH2 + HBr}$$ So A is ethene. Ozonolysis then cleaves that double bond into carbonyl compounds, and...
2220732 marksHideAnswer
How would you prepare methane from i) chloromethane ii) sodium ethanoate?
(i) From chloromethane (reduction by nascent hydrogen, Zn/HCl couple): $\ce{CH3Cl + 2[H] -[Zn/Cu] CH4 + HCl}$ (ii) From sodium ethanoate (decarboxylation by heating with soda lime, NaOH + CaO): $$\ce{CH3COONa + NaOH -[CaO][\Delta] CH4 ...
(i) From chloromethane (reduction by nascent hydrogen, Zn/HCl couple): $$\ce{CH3Cl + 2[H] -[Zn/Cu] CH4 + HCl}$$ (ii) From sodium ethanoate (decarboxylation by heating with soda lime, NaOH + CaO): $$\ce{CH3COONa + NaOH -[CaO][\Delta] CH4 ...
2320735 marksHideAnswer
What property did Mendeleev use to classify the elements in his periodic table? Point out the anomalies in Mendeleev's periodic table.
Basis: Mendeleev classified the elements in order of their increasing atomic mass, grouping together elements with similar properties (his law: properties are a periodic function of atomic mass).

Anomalies (defects):
1. Position of hydrogen was not fixed (it resembles both Group I and Group VII).
2. Anomalous pairs: in some places a heavier element was placed before a lighter one to keep similar elements together, e.g. Ar (40) before K (39), Co before Ni, Te before I.
3. Isotopes had no separate place (different masses, same element).
4. Grouping of dissimilar elements together in sub-groups (e.g. coinage metals with alkali metals).
5. The position of the lanthanides and actinides could not be accommodated.
2420735 marksHideAnswer
Balance the following equation by the oxidation number or ion-electron method and point out the oxidant and reductant: Zn + NaNO3 + NaOH -> Na2ZnO2 + NH3 + H2O.
Zn goes $0\to+2$ (loses $2e^-$ ); N goes $+5\to-3$ (gains $8e^-$ ). To balance electrons, 4 Zn ( $8e^-$ lost) per 1 N ( $8e^-$ gained): $\ce{4Zn + NaNO3 + 7NaOH - 4Na2ZnO2 + NH3 + 2H2O}$ Check: Na $$ \begin{aligned} 1+7 &= 8 \ &= 4\times2...
Zn goes $0\to+2$ (loses $2e^-$); N goes $+5\to-3$ (gains $8e^-$). To balance electrons, 4 Zn ($8e^-$ lost) per 1 N ($8e^-$ gained): $$\ce{4Zn + NaNO3 + 7NaOH - 4Na2ZnO2 + NH3 + 2H2O}$$ Check: Na $$ \begin{aligned} 1+7 &= 8 \ &= 4\times2...
2520735 marksHideAnswer
How is ethene prepared in the laboratory? What happens when ethene is passed through Baeyer's reagent?
Laboratory preparation of ethene (dehydration of ethanol by heating with excess conc. $\ce{H2SO4}$ at ~ $170\,^\circ\text{C}$ ): $\ce{C2H5OH -[conc.\ H2SO4][170\,^\circ C] C2H4(g) + H2O}$ The gas is washed and collected over water. With ...
Laboratory preparation of ethene (dehydration of ethanol by heating with excess conc. $\ce{H2SO4}$ at ~$170,^\circ\text{C}$): $$\ce{C2H5OH -[conc.\ H2SO4][170,^\circ C] C2H4(g) + H2O}$$ The gas is washed and collected over water. With ...
2620735 marksHideAnswer
Starting from sulphur, how would you obtain sulphuric acid (reactions only)? How does conc. H2SO4 react with i) sugar ii) P4? Why is conc. sulphuric acid diluted by adding acid to water and not water to acid?
Sulphur to sulphuric acid (contact process, reactions only): $\begin{aligned} \ce{S + O2 - SO2} \\ \ce{2SO2 + O2 <=[V2O5] 2SO3} \\ \ce{SO3 + H2SO4 - H2S2O7} \\ \ce{H2S2O7 + H2O - 2H2SO4} \end{aligned}$ i) With sugar (dehydration to b...
Sulphur to sulphuric acid (contact process, reactions only): $$ \begin{aligned} \ce{S + O2 - SO2} \ \ce{2SO2 + O2 <=[V2O5] 2SO3} \ \ce{SO3 + H2SO4 - H2S2O7} \ \ce{H2S2O7 + H2O - 2H2SO4} \end{aligned} $$ i) With sugar (dehydration to b...
2720735 marksHideAnswer
Differentiate between i) calcination and roasting ii) carbon-reduction process and thermite process.
i) Calcination vs roasting:
- Calcination is heating an ore strongly in limited/absence of air to expel moisture and volatile matter and to convert carbonates/hydroxides to oxides, e.g.
$\ce{CaCO3 ->[\Delta] CaO + CO2}$
.
- Roasting is heating an ore strongly in a current of air, mainly for sulphide ores, converting them to oxides with release of $\ce{SO2}$ , e.g.
$\ce{2ZnS + 3O2 -> 2ZnO + 2SO2}$
.

ii) Carbon-reduction vs thermite process:
- Carbon reduction uses coke to reduce a metal oxide, e.g.
$\ce{ZnO + C -> Zn + CO}$
; suitable for moderately reactive metals.
- Thermite (alumino-thermite) process uses aluminium as the reducing agent for oxides not easily reduced by carbon, e.g.
$\ce{Fe2O3 + 2Al -> 2Fe + Al2O3}$
; highly exothermic, used to weld rails.
i) Calcination vs roasting:
- Calcination is heating an ore strongly in limited/absence of air to expel moisture and volatile matter and to convert carbonates/hydroxides to oxides, e.g.
$$ \ce{CaCO3 ->[\Delta] CaO + CO2} $$

.
- Roasting is heating an ore strongly in a current of air, mainly for sulphide ores, converting them to oxides with release of $\ce{SO2}$, e.g.
$$ \ce{2ZnS + 3O2 -> 2ZnO + 2SO2} $$

.

ii) Carbon-reduction vs thermite process:
- Carbon reduction uses coke to reduce a metal oxide, e.g.
$$ \ce{ZnO + C -> Zn + CO} $$

; suitable for moderately reactive metals.
- Thermite (alumino-thermite) process uses aluminium as the reducing agent for oxides not easily reduced by carbon, e.g.
$$ \ce{Fe2O3 + 2Al -> 2Fe + Al2O3} $$

; highly exothermic, used to weld rails.
2820735 marksHideAnswer
Describe the preparation of bromine from carnallite. What happens when bromine is passed through hot and conc. NaOH?
Preparation from carnallite: the bromide-rich bittern (mother liquor from carnallite $\ce{KCl.MgCl2.6H2O}$ ) is acidified and warmed, and chlorine is passed through it; chlorine displaces bromine, which is swept out by air/steam and condensed: $\ce{MgBr2 + Cl2 -> MgCl2 + Br2(g)}$

Bromine with hot conc. NaOH (disproportionation to bromide and bromate): $\ce{3Br2 + 6NaOH(hot,\ conc) -> 5NaBr + NaBrO3 + 3H2O}$ (With cold dilute NaOH it gives $\ce{NaBr}$ + $\ce{NaBrO}$ instead.)
Preparation from carnallite: the bromide-rich bittern (mother liquor from carnallite $\ce{KCl.MgCl2.6H2O}$) is acidified and warmed, and chlorine is passed through it; chlorine displaces bromine, which is swept out by air/steam and condensed: $$\ce{MgBr2 + Cl2 -> MgCl2 + Br2(g)}$$

Bromine with hot conc. NaOH (disproportionation to bromide and bromate): $$\ce{3Br2 + 6NaOH(hot,\ conc) -> 5NaBr + NaBrO3 + 3H2O}$$ (With cold dilute NaOH it gives $\ce{NaBr}$ + $\ce{NaBrO}$ instead.)
2920735 marksNumericalHideAnswer
For the reaction 2NaHCO3 --(heat)--> Na2CO3 + H2O + CO2, 2.5 g of a sample of NaHCO3 when strongly heated gives 310 cc of CO2 at 27 degree C and 760 mmHg. i) Calculate the percentage purity of the sample. ii) How many moles of water are produced? iii) What mass of pure HCl is required to neutralize the Na2CO3 produced?
$\begin{aligned} \ce{2NaHCO3 ->[\Delta] Na2CO3 + H2O + CO2} \\ n_{\ce{CO2}} &= \dfrac{PV}{RT} \\ &= \dfrac{1\times0.310}{0.0821\times300} \\ &= 0.01259\ \text{mol} \end{aligned}$
i) Purity:
$\ce{2NaHCO3 -> 1CO2}$
, so
$\begin{aligned} \text{pure }\ce{NaHCO3} &= 2\times0.01259 \\ &= 0.02517\ \text{mol} \\ &= 0.02517\times84 \\ &= 2.115\ \text{g} \\ \%\ \text{purity} &= \dfrac{2.115}{2.5}\times100 \\ &= \mathbf{84.6\%} \end{aligned}$
ii) Water:

$1\ \ce{H2O}$ per $1\ \ce{CO2}$ , so $n_{\ce{H2O}}=\mathbf{0.0126\ mol}$ .

iii) HCl to neutralise $\ce{Na2CO3}$ :

$n_{\ce{Na2CO3}}=0.01259$ mol.
$\begin{aligned} \ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2} \\ n_{\ce{HCl}} &= 2\times0.01259 \\ &= 0.02517\ \text{mol} \\ \text{mass} &= 0.02517\times36.5 \\ &= \mathbf{0.919\ g} \end{aligned}$
$$ \begin{aligned} \ce{2NaHCO3 ->[\Delta] Na2CO3 + H2O + CO2} \ n_{\ce{CO2}} &= \dfrac{PV}{RT} \ &= \dfrac{1\times0.310}{0.0821\times300} \ &= 0.01259\ \text{mol} \end{aligned} $$

i) Purity:

$$ \ce{2NaHCO3 -> 1CO2} $$

, so

$$ \begin{aligned} \text{pure }\ce{NaHCO3} &= 2\times0.01259 \ &= 0.02517\ \text{mol} \ &= 0.02517\times84 \ &= 2.115\ \text{g} \ %\ \text{purity} &= \dfrac{2.115}{2.5}\times100 \ &= \mathbf{84.6%} \end{aligned} $$

ii) Water:

$1\ \ce{H2O}$ per $1\ \ce{CO2}$, so $n_{\ce{H2O}}=\mathbf{0.0126\ mol}$.

iii) HCl to neutralise $\ce{Na2CO3}$:

$n_{\ce{Na2CO3}}=0.01259$ mol.

$$ \begin{aligned} \ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2} \ n_{\ce{HCl}} &= 2\times0.01259 \ &= 0.02517\ \text{mol} \ \text{mass} &= 0.02517\times36.5 \ &= \mathbf{0.919\ g} \end{aligned} $$
30207310 marksNumericalHideAnswer
State Charle's law. How did this law lead to the development of the absolute temperature scale? Plot the temperature-volume relationship indicating absolute zero. A saturated hydrocarbon (CnH2n+2) diffuses through a porous membrane twice as fast as sulphur dioxide. Determine the molecular formula of the hydrocarbon.
Charles's law: at constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature: $V\propto T$ , i.e. $\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$ .

Absolute scale: experimentally the volume falls by $\tfrac{1}{273}$ of its $0\,^\circ\text{C}$ value per degree of cooling. Extending the $V$ - $t$ line, $V$ would become zero at $-273\,^\circ\text{C}$ ; a lower temperature (negative volume) is impossible, so this is absolute zero and the Kelvin scale starts there: $T(\text{K})=t(^\circ\text{C})+273$ .

Plot: a straight line of $V$ (y-axis) against $t\ (^\circ\text{C})$ (x-axis) that, when extrapolated backwards, meets the temperature axis at $-273\,^\circ\text{C}$ (where $V=0$ ), which marks absolute zero.

Molecular formula (Graham's law):
$\begin{aligned} \dfrac{r_{HC}}{r_{\ce{SO2}}} &= 2 \\ &= \sqrt{\dfrac{M_{\ce{SO2}}}{M_{HC}}} \\ 4 &= \dfrac{64}{M_{HC}} \\ M_{HC} &= 16 \\ 14n+2 &= 16 \\ &\Rightarrow n = 1 \end{aligned}$
The hydrocarbon is methane, $\ce{CH4}$ .
Charles's law: at constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature: $V\propto T$, i.e. $\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$.

Absolute scale: experimentally the volume falls by $\tfrac{1}{273}$ of its $0,^\circ\text{C}$ value per degree of cooling. Extending the $V$-$t$ line, $V$ would become zero at $-273,^\circ\text{C}$; a lower temperature (negative volume) is impossible, so this is absolute zero and the Kelvin scale starts there: $T(\text{K})=t(^\circ\text{C})+273$.

Plot: a straight line of $V$ (y-axis) against $t\ (^\circ\text{C})$ (x-axis) that, when extrapolated backwards, meets the temperature axis at $-273,^\circ\text{C}$ (where $V=0$), which marks absolute zero.

Molecular formula (Graham's law):

$$ \begin{aligned} \dfrac{r_{HC}}{r_{\ce{SO2}}} &= 2 \ &= \sqrt{\dfrac{M_{\ce{SO2}}}{M_{HC}}} \ 4 &= \dfrac{64}{M_{HC}} \ M_{HC} &= 16 \ 14n+2 &= 16 \ &\Rightarrow n = 1 \end{aligned} $$

The hydrocarbon is methane, $\ce{CH4}$.
31207310 marksHideAnswer
Sketch a well-labelled diagram for the manufacture of caustic soda (NaOH) by Castner-Kellner's process. Explain the principle and procedure involved. What happens when caustic soda is i) heated with sulphur ii) treated with zinc? How would you convert caustic soda into sodium silicate?
Castner-Kellner process: brine is electrolysed with a flowing mercury cathode and graphite anodes. $$ \begin{aligned} \text{Anode: }\ce{2Cl- - Cl2(g) + 2e-} \ \text{Cathode: }\ce{Na+ + e- - Na}\ (\to \text{sodium amalgam}) \end{aligned}...
32207310 marksHideAnswer
Write the principle and process involved along with a self-explanatory diagram for the manufacture of nitric acid by catalytic oxidation of ammonia. How does concentrated nitric acid react with i) iron ii) iodine? Why is conc. nitric acid stored in a dark bottle?
Ostwald's process (catalytic oxidation of ammonia) - principle and process:
$\begin{aligned} \ce{4NH3(g) + 5O2(g) ->[Pt/Rh][~800\,^\circ C] 4NO(g) + 6H2O(g)} \\ \ce{2NO(g) + O2(g) -> 2NO2(g)} \\ \ce{4NO2(g) + O2(g) + 2H2O(l) -> 4HNO3(aq)} \end{aligned}$
A $1:8$ ammonia-air mixture is passed over Pt/Rh gauze (~ $800\,^\circ\text{C}$ ); the $\ce{NO}$ is cooled, oxidised to $\ce{NO2}$ by more air, and absorbed in water in a tower to give nitric acid.

i) Conc. $\ce{HNO3}$ + iron: cold conc. $\ce{HNO3}$ makes iron passive, forming a thin protective oxide layer that stops further reaction (so it is transported in iron/steel tankers).

ii) Conc. $\ce{HNO3}$ + iodine (oxidises $\ce{I2}$ to iodic acid): $\ce{I2 + 10HNO3 -> 2HIO3 + 10NO2(g) + 4H2O}$

Dark bottle: nitric acid is decomposed by light (and heat) into brown $\ce{NO2}$ , which colours it yellow: $\ce{4HNO3 ->[light] 4NO2 + O2 + 2H2O}$ A dark (amber) bottle keeps out light and prevents this decomposition.
Ostwald's process (catalytic oxidation of ammonia) - principle and process: $$ \begin{aligned} \ce{4NH3(g) + 5O2(g) ->[Pt/Rh][~800,^\circ C] 4NO(g) + 6H2O(g)} \ \ce{2NO(g) + O2(g) -> 2NO2(g)} \ \ce{4NO2(g) + O2(g) + 2H2O(l) -> 4HNO3(aq)} \end{aligned} $$ A $1:8$ ammonia-air mixture is passed over Pt/Rh gauze (~$800,^\circ\text{C}$); the $\ce{NO}$ is cooled, oxidised to $\ce{NO2}$ by more air, and absorbed in water in a tower to give nitric acid.

i) Conc. $\ce{HNO3}$ + iron: cold conc. $\ce{HNO3}$ makes iron passive, forming a thin protective oxide layer that stops further reaction (so it is transported in iron/steel tankers).

ii) Conc. $\ce{HNO3}$ + iodine (oxidises $\ce{I2}$ to iodic acid): $$\ce{I2 + 10HNO3 -> 2HIO3 + 10NO2(g) + 4H2O}$$

Dark bottle: nitric acid is decomposed by light (and heat) into brown $\ce{NO2}$, which colours it yellow: $$\ce{4HNO3 ->[light] 4NO2 + O2 + 2H2O}$$ A dark (amber) bottle keeps out light and prevents this decomposition.
33207310 marksHideAnswer
Write short notes on any two: i) Rutherford atomic model ii) Le-Chatelier's principle and its applications iii) Characteristics of homologous series iv) Lassaigne's test for nitrogen.
i) Rutherford atomic model: from the $\alpha$ -scattering experiment, Rutherford proposed that an atom has a tiny, dense, positively charged nucleus carrying almost all its mass, with electrons revolving around it in the mostly empty space; the atom is electrically neutral. Its defect: it could not explain atomic stability or line spectra.

ii) Le-Chatelier's principle: if a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to undo the change. Applications: in Haber's process ( $\ce{N2 + 3H2 <=> 2NH3}$ ) high pressure and low temperature increase $\ce{NH3}$ yield; in the contact process high pressure/excess $\ce{O2}$ increase $\ce{SO3}$ .

iii) Characteristics of homologous series: same general formula and same functional group; successive members differ by $\ce{-CH2-}$ ( $14\ \text{u}$ ); gradual gradation of physical properties with molar mass; similar chemical properties; all made by similar general methods.

iv) Lassaigne's test for nitrogen: the compound is fused with sodium to give $\ce{NaCN}$ ; the extract is boiled with $\ce{FeSO4}$ and acidified with $\ce{HCl}$ ; a Prussian-blue colour confirms nitrogen:
$\ce{6NaCN + FeSO4 -> Na4[Fe(CN)6] + Na2SO4}$
;
$\ce{3Na4[Fe(CN)6] + 4FeCl3 -> Fe4[Fe(CN)6]3 + 12NaCl}$
.
i) Rutherford atomic model: from the $\alpha$-scattering experiment, Rutherford proposed that an atom has a tiny, dense, positively charged nucleus carrying almost all its mass, with electrons revolving around it in the mostly empty space; the atom is electrically neutral. Its defect: it could not explain atomic stability or line spectra.

ii) Le-Chatelier's principle: if a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to undo the change. Applications: in Haber's process ($\ce{N2 + 3H2 <=> 2NH3}$) high pressure and low temperature increase $\ce{NH3}$ yield; in the contact process high pressure/excess $\ce{O2}$ increase $\ce{SO3}$.

iii) Characteristics of homologous series: same general formula and same functional group; successive members differ by $\ce{-CH2-}$ ($14\ \text{u}$); gradual gradation of physical properties with molar mass; similar chemical properties; all made by similar general methods.

iv) Lassaigne's test for nitrogen: the compound is fused with sodium to give $\ce{NaCN}$; the extract is boiled with $\ce{FeSO4}$ and acidified with $\ce{HCl}$; a Prussian-blue colour confirms nitrogen:

$$ \ce{6NaCN + FeSO4 -> Na4[Fe(CN)6] + Na2SO4} $$

;

$$ \ce{3Na4[Fe(CN)6] + 4FeCl3 -> Fe4[Fe(CN)6]3 + 12NaCl} $$

.

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