CBSE Class 12 Chemistry 2023 Delhi Set 1 Solved Paper

Question : 33

Total: 35

SECTION - E

(a) (i) Why is boiling point of 1MNaCl solution more than that of 1M glucose solution?
(ii) A nonvolatile solute ' X ' (molar mass =50 g mol−1 ) when dissolved in 78g of benzene reduced its vapour pressure to 90%.
Calculate the mass of X dissolved in the solution.
(iii) Calculate the boiling point elevation for a solution prepared by adding 10g of MgCl2 to 200g of water assuming MgCl2 is completely dissociated.
(Kb. for Water =0.512 K kg mol−1, Molar mass MgCl2=95g 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 2.32×10−2g of K2SO4 in 2L of solution at 25°C, assuming that K2SO4 is completely dissociated.
(R=0.082Latm K−1 mol−1., Molar mass K2SO4 =174 g mol−1 )
(iii) When 25.6g of sulphur was dissolved in 1000g of benzene, the freezing point lowered by 0.512K. Calculate the formula of sulphur (Sx).
(Kf. for benzene =5.12K kg mol−1, Atomic mass of Sulphur =32 g mol−1 )

Solution: 👈: Video Solution

(a) (i) 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 will required high temperature to boil while glucose need low temperature to dissociateits bonding.
(ii)

P°−PS

w×M

m×w

100−90

w×78

50×78

90×w=10×50
w=

10×50

=5.55 grams
(ii) Tb=i×kb×m(MgCl2=3. ions )i=3
Tb=

3×0.512×10

95×0.2

(200 gm=0.2 Kg)
=

5.36

=0.80
So elevation is boiling point =273+0.80 =273.80K
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 cocentration

Theoretical number of particles

∵ It is considered that degree of dissociation of ethanoicacid in benzene is 1
2CH3−COOH→(CH3−COOH)2
So, i=

=0.5
(ii) K2SO4→2K++SO4‌2−
ions produced are =3
i=3
x=iCRT
=3×

0.0232

174

×0.5×0.0821×298
=

0.851

174

=0.00489
=4.89×10−3 atm
(iii) ΔTf=

Kf×m×1000

MB×WA

[∵MB=Sx].
0.512=

5.12×25.6×1000

~Sx×1000

Sx=

5.12×25.6

0.512

Sx=256
x×32=256
x=

256

=8
So, the required formula (Sx)
x=8
or S8.

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