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✓ Board-verified
Chemistry · 2023 · 5 marks
CBSE 2023 · Region 5 · Set 1 · Q33
(a) (i) Why is boiling point of 1 M NaCl solution more than that of 1 M glucose solution? (ii) A non-volatile solute 'X' (molar mass =50 g mol^-1 )…
(i)
Why is boiling point of $\displaystyle 1$ M NaCl solution more than that of $\displaystyle 1$ M glucose solution ?
(ii)
A non-volatile solute 'X' (molar mass $\displaystyle =50 \mathrm{~g} \mathrm{~mol}^{-1}$ ) when dissolved in 78g of benzene reduced its vapour pressure to $\displaystyle 90$%. Calculate the mass of X dissolved in the solution.
(iii)
Calculate the boiling point elevation for a solution prepared by adding 10g of $\displaystyle \mathrm{MgCl}_{2}$ to 200g of water assuming $\displaystyle \mathrm{MgCl}_{2}$ is completely dissociated. ( $\displaystyle \mathrm{K}_{\mathrm{b}}$ for Water $\displaystyle =0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, Molar mass $\displaystyle \mathrm{MgCl}_{2}=95 \mathrm{~g} \mathrm{~mol}^{-1}$ )
(i)
Why is the value of Van't Hoff factor for ethanoic acid in benzene close to $\displaystyle 0.5$ ?
(ii)
Determine the osmotic pressure of a solution prepared by dissolving $\displaystyle 2.32 \times 10^{-2} \mathrm{~g}$ of $\displaystyle \mathrm{K}_{2} \mathrm{SO}_{4}$ in 2L of solution at $\displaystyle 25$ °C, assuming that $\displaystyle \mathrm{K}_{2} \mathrm{SO}_{4}$ is completely dissociated. ( $\displaystyle \mathrm{R}=0.082 \mathrm{~L}$ atm $\displaystyle \mathrm{K}^{-1} \mathrm{~mol}^{-1}$, Molar mass $\displaystyle \mathrm{K}_{2} \mathrm{SO}_{4}=174 \mathrm{~g} \mathrm{~mol}^{-1}$ )
(iii)
When 25.6g of Sulphur was dissolved in 1000g of benzene, the freezing point lowered by $\displaystyle 0.512$ K. Calculate the formula of Sulphur $\displaystyle \left(\mathrm{S}_{\mathrm{x}}\right)$. $\displaystyle \left(\mathrm{K}_{\mathrm{f}}\right.$ for benzene $\displaystyle =5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, Atomic mass of Sulphur = 32g $\displaystyle \mathrm{mol}^{-1}$ )
Marking-scheme solution
(a)(i)
Dissociation of NaCl/more number of particles in NaCl solution/Value of 'i' for NaCl is greater than that of glucose.
(ii)
\[\begin{aligned}
& \frac{\mathrm{p}^{o}-\mathrm{p}}{\mathrm{p}^{o}}=\mathrm{x}_{2} \\
& \quad=\frac{n_{2}}{n_{2}+n_{1}} \\
& \mathrm{p}=0.9 \mathrm{p}^{0} \\
& \frac{\mathrm{p}^{o}-0.9 \mathrm{p}^{o}}{\mathrm{p}^{o}}=\frac{\mathrm{w} / 50}{\mathrm{w} / 50+1} \\
& \quad 0 \cdot 1\left(\frac{\mathrm{w}}{50}+1\right)=\frac{\mathrm{w}}{50}
\end{aligned}
\] (or 5g if dilute solution is considered) .
(iii)
\[\begin{aligned}
\Delta \mathrm{T}_{\mathrm{b}} & =\mathrm{i} \mathrm{~K}_{\mathrm{b}} \mathrm{~m} \\
\mathrm{i} & =3 \\
\Delta \mathrm{~T}_{\mathrm{b}} & =3 \times 0.512 \times \frac{10}{95} \times \frac{1000}{200} \\
\Delta \mathrm{~T}_{\mathrm{b}} & =0.81 \mathrm{~K} \text { or }{ }^{\circ} \mathrm{C}
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.