CBSE Class 12 Chemistry 2023 Delhi Set 1 Solved Paper

Question : 33 of 35

Marks: +1, -0

SECTION - E

(a) (i) Why is boiling point of

1 {M} {NaCl}

solution more than that of

1 {M}

glucose solution?
(ii) A nonvolatile solute '

{X}

' (molar mass

={5 0}

{g} \ {\mol}^{-1}

) when dissolved in

78 {g}

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

10 {g}

{MgCl}_2

200 {g}

of water assuming

{MgCl}_2

is completely dissociated.

({K}_{{b}}.

for Water

=0.512 \ { K} \ { kg} \ { mol}^{-1}

, Molar mass

{MgCl}_2=95 { g} \ { 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^{-2} {g}

{K}_2 {SO}_4

2 {L}

of solution at

25°{C}

, assuming that

{K}_2 {SO}_4

is completely dissociated.

({R}=0.082 {L} { atm} \ { K}^{-1} \ { mol}^{-1}.

, Molar mass

{K}_2 {SO}_4

=174 \ { g} \ { mol}^{-1}

)
(iii) When

25.6 {g}

of sulphur was dissolved in

1000 {g}

of benzene, the freezing point lowered by

0.512 {K}

. Calculate the formula of sulphur

({S}_{{x}})

({K}_{{f}}.

for benzene

=5.12 { K}\ {kg}\ { mol}^{-1}

, Atomic mass of Sulphur

=32 \ { g} \ { mol}^{-1}

)

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}°-{P}_{{S}}}/{{P}_{{s}}}={{n}}/{{N}}={{w} \× {M}}/{{m} \×{w}}

{100-90}/{90}={w \×78}/{50 \×78}

{10}/{90}={w}/{50}

90 \×w=10 \×50

{w}={10 \×50}/{90}=5.55

grams
(ii)

{T}_{{b}}={i} \×{k}_{{b}}\×{m}({MgCl}_2=3.

ions

) {i}=3

{T}_{{b}}={3 \× 0.512 \×10}/{95 \×0.2}

(200 \ {gm}=0.2 \ {Kg})

={5.36}/{19}=0.80

So elevation is boiling point

=273+0.80

=273.80 {K}

OR
(b) (i) Van't Hoff factor is used indetermination of colligative property. It is also called Abnormal colligative property.

i={\text "actual no. of particle or cocentration "}/{\text "Theoretical number of particles "}

\∵

It is considered that degree of dissociation of ethanoicacid in benzene is 1

2 {CH}_3-{COOH} \→({CH}_3-{COOH})_2

So,

i={1}/{2}=0.5

(ii)

{K}_2 {SO}_4 \→2 {K}^{+}+{SO}_4{ }^{2-}

ions produced are

=3

i=3

{x}=i {CRT}

=3 \×{0.0232}/{174} \×0.5 \×0.0821 \×298

={0.851}/{174}=0.00489

=4.89 \×10^{-3} \ { atm}

(iii)

\Δ{T}_{{f}}={{K}_{{f}} \×{m} \×1000}/{{M}_{{B}} \×{W}_{{A}}}

\[\∵{M}_{{B}}={S}_{{x}}]

0.512={5.12 \×25.6 \×1000}/{{~S}_{{x}} \×1000}

S_x={5.12 \×25.6}/{0.512}

S_x=256

x \×32=256

x={256}/{32}=8

So, the required formula

({S}_{{x}})

x=8

{S}_8

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