(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 of K$_2$SO$_4$ in 2 L of solution at 25°C, assuming that K$_2$SO$_4$ is completely dissociated. (R = 0.082 L atm mol$^{-1}$ K$^{-1}$, molar mass of 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$^{-1}$)

(i) Ethanoic acid dimerises almost completely in benzene; for full dimerisation i = 1 − α/2 → 0.5. (ii) π ≈ 4.89 × 10⁻³ atm (iii) S₈

Solutions — CBSE Class 12 · 2023 Board Question with Solution

(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 of K

_{2}

_{4}

in 2 L of solution at 25°C, assuming that K

_{2}

_{4}

is completely dissociated. (R = 0.082 L atm mol

^{- 1}

^{- 1}

, molar mass of K

_{2}

_{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

(i) In benzene, ethanoic acid molecules pair up through hydrogen bonds into dimers. For dimerisation

i = 1 - \frac{α}{2}

; with nearly complete association (

α \to 1

i \to 0.5

— hence the observed value close to 0.5.
(ii) Moles of K

_{2}

_{4}

n = \frac{2.32 \times 1 0 ^{- 2}}{174} = 1.333 \times 1 0^{- 4} mol

Complete dissociation (2K

^{+}

+ SO

_{4}^{2 -}

) gives

i = 3

π = \frac{i n R T}{V} = \frac{3 \times 1.333 \times 1 0 ^{- 4} \times 0.082 \times 298}{2}

π = 4.89 \times 1 0^{- 3} atm

(*) The booklet's OCR garbles the mass; the CBSE 2023 paper value

2.32 \times 1 0^{- 2}

g is used.
(iii) Molality:

m = \frac{0.512}{5.12} = 0.100 mol kg^{- 1}

Moles in 1000 g benzene = 0.100 mol, so

M = \frac{25.6}{0.100} = 256 g mol^{- 1}, x = \frac{256}{32} = 8

The formula is S_{8}

CBSE Class 12 Board · 2023 · Subjective

CBSE Class 12 · 2023 · SolutionsLong Answer

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