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Q.Phenol in water partially associates to give dimer molecule. 0.94 g C6H5OHC_6H_5OH dissolved in 50 g water freezes at −0.21°C-0.21°C. Calculate van't Hoff factor (molal freezing point of water is 1.86°C1.86°C) of the solution. OR H2SO4H_2SO_4 used in lead storage cell is 38% by weight and its density is 1.30 gcm−3^{-3}. Calculate the molality of the H2SO4H_2SO_4 solution.

Manipur CohsemCOHSEM Manipur Higher Secondary Board 2024Subjective· 3mImportance★★★★★
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This question offers a choice between a van't Hoff factor calculation for phenol's association (primary) or the molality of battery-acid H₂SO₄ (the alternative). Both are answered below.

(Primary) — van't Hoff factor for phenol association:

Molality of phenol (without considering association): moles of C6H5OHC_6H_5OH (M=94 g/molM=94\ g/mol) =0.9494=0.01 mol= \dfrac{0.94}{94} = 0.01\ mol; mass of water =50 g=0.05 kg= 50\ g = 0.05\ kg.

m=0.010.05=0.2 mol kg−1m = \frac{0.01}{0.05} = 0.2\ mol\,kg^{-1}

Calculated (normal) freezing point depression: ΔTf(calc)=Kf×m=1.86×0.2=0.372°C\Delta T_f(\text{calc}) = K_f \times m = 1.86 \times 0.2 = 0.372°C

Observed freezing point depression: ΔTf(obs)=0−(−0.21)=0.21°C\Delta T_f(\text{obs}) = 0 - (-0.21) = 0.21°C

i=ΔTf(obs)ΔTf(calc)=0.210.372=0.565i = \frac{\Delta T_f(\text{obs})}{\Delta T_f(\text{calc})} = \frac{0.21}{0.372} = 0.565

Since i<1i < 1, this confirms association (phenol molecules partially dimerising in water via hydrogen bonding), consistent with the observation stated in the question.

OR — Molality of the H2SO4H_2SO_4 solution:

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