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CBSE Class 12 Chemistry 2023 Delhi Set 1 Solved Paper

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Question : 33 of 35
Marks: +1, -0
SECTION - E
(a) (i) Why is boiling point of 1MNaCl1 \text{M} \mathrm{NaCl} solution more than that of 1M1 \text{M} glucose solution?
(ii) A nonvolatile solute ' XX ' (molar mass =50=5 0 g mol−1\text{g mol}^{-1} ) when dissolved in 78g78 \text{g} of benzene reduced its vapour pressure to 90%90\%.
Calculate the mass of XX dissolved in the solution.
(iii) Calculate the boiling point elevation for a solution prepared by adding 10g10 \text{g} of MgCl2\mathrm{MgCl}_2 to 200g200 \text{g} of water assuming MgCl2\mathrm{MgCl}_2 is completely dissociated.
(Kb.(K_{b}. for Water =0.512 K kg mol−1=0.512\ \mathrm{K}\ \mathrm{kg}\ \mathrm{mol}^{-1}, Molar mass MgCl2=95 g mol−1\mathrm{MgCl}_2=95\ \mathrm{g}\ \mathrm{mol}^{-1} )
OR
(b) (i) Why is the value of Van't Hoff factor for ethanoic acid in benzene close to 0.5 ?
(ii) Determine the osmotic pressure of a solution prepared by dissolving 2.32×10−2g2.32 \times 10^{-2} \text{g} of K2SO4\mathrm{K}_2\mathrm{SO}_4 in 2L2 \text{L} of solution at 25∘C25^{\circ}\mathrm{C}, assuming that K2SO4\mathrm{K}_2\mathrm{SO}_4 is completely dissociated.
(R=0.082 L atm K−1 mol−1.(R=0.082\ \mathrm{L}\ \mathrm{atm}\ \mathrm{K}^{-1}\ \mathrm{mol}^{-1}., Molar mass K2SO4\mathrm{K}_2\mathrm{SO}_4 =174 g mol−1=174\ \mathrm{g}\ \mathrm{mol}^{-1} )
(iii) When 25.6g25.6 \text{g} of sulphur was dissolved in 1000g1000 \text{g} of benzene, the freezing point lowered by 0.512 K0.512\ \mathrm{K}. Calculate the formula of sulphur (Sx)(S_{x}).
(Kf.(K_{f}. for benzene =5.12 K kg mol−1=5.12\ \mathrm{K}\ \mathrm{kg}\ \mathrm{mol}^{-1}, Atomic mass of Sulphur =32 g mol−1=32\ \mathrm{g}\ \mathrm{mol}^{-1} )
Solution:     👈: Video Solution
(a) (i) NaCl\mathrm{NaCl} is having ionic bonding between sodium and chloride atoms which is strong bonding while glucose having covalent bonding which is weak in nature. NaCl\mathrm{NaCl} will required high temperature to boil while glucose need low temperature to dissociateits bonding.
(ii) P∘−PSPs=nN=w×Mm×w\frac{P^{\circ}-P_{S}}{P_{s}}=\frac{n}{N}=\frac{w \times M}{m \times w}
100−9090=w×7850×78\frac{100-90}{90}=\frac{w \times 78}{50 \times 78}
1090=w50\frac{10}{90}=\frac{w}{50}
90×w=10×5090 \times w = 10 \times 50
w=10×5090=5.55w=\frac{10 \times 50}{90}=5.55 grams
(ii) Tb=i×kb×m(MgCl2=3.T_{b}=i \times k_{b} \times m (\mathrm{MgCl}_2=3. ions )i=3) i=3
Tb=3×0.512×1095×0.2T_{b}=\frac{3 \times 0.512 \times 10}{95 \times 0.2} (200 gm=0.2 Kg)(200\ \mathrm{gm}=0.2\ \mathrm{Kg})
=5.3619=0.80=\frac{5.36}{19}=0.80
So elevation is boiling point =273+0.80=273+0.80 =273.80 K=273.80\ \mathrm{K}
OR
(b) (i) Van't Hoff factor is used indetermination of colligative property. It is also called Abnormal colligative property.
i=actual no. of particle or cocentrationTheoretical number of particlesi=\frac{\text{actual no. of particle or cocentration}}{\text{Theoretical number of particles}}
∵\because It is considered that degree of dissociation of ethanoicacid in benzene is 1
2CH3−COOH→(CH3−COOH)22 \mathrm{CH}_3-\mathrm{COOH} \rightarrow (\mathrm{CH}_3-\mathrm{COOH})_2
So, i=12=0.5i=\frac{1}{2}=0.5
(ii) K2SO4→2K++SO42−\mathrm{K}_2\mathrm{SO}_4 \rightarrow 2 \mathrm{K}^{+} + \mathrm{SO}_4^{2-}
ions produced are =3=3
i=3i=3
x=iCRTx = i CRT
=3×0.0232174×0.5×0.0821×298=3 \times \frac{0.0232}{174} \times 0.5 \times 0.0821 \times 298
=0.851174=0.00489=\frac{0.851}{174}=0.00489
=4.89×10−3 atm=4.89 \times 10^{-3}\ \mathrm{atm}
(iii) ΔTf=Kf×m×1000MB×WA\Delta T_{f}=\frac{K_{f} \times m \times 1000}{M_{B} \times W_{A}} [∵MB=Sx][\because M_{B}=S_{x}].
0.512=5.12×25.6×1000Sx×10000.512=\frac{5.12 \times 25.6 \times 1000}{S_{x} \times 1000}
Sx=5.12×25.60.512S_x=\frac{5.12 \times 25.6}{0.512}
Sx=256S_x=256
x×32=256x \times 32 = 256
x=25632=8x=\frac{256}{32}=8
So, the required formula (Sx)(S_{x})
x=8x=8
or S8\mathrm{S}_8.
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