When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be $-0.93^{\circ} \mathrm{C}$ $[\mathrm{K}_{\mathrm{f}}(\mathrm{H}_{2} \mathrm{O})=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}]$. The number (n) of benzoic acid molecules associated (assuming 100% association) is ____

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Solutions

When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be

- 0.9 3^{\circ} C

[K_{f} (H_{2} O) = 1.86 K kg mol^{- 1}]

. The number (n) of benzoic acid molecules associated (assuming 100% association) is ____

Answer: 2

Solution

The molality of the solution is

m = \frac{12.2/122}{0.100} = 1.00 mol k g^{- 1}

. From

Δ T_{f} = i K_{f} m

i = \frac{0.93}{1.86 \times 1.00} = 0.5

A van't Hoff factor below one means the solute associates. If

n

molecules combine into one particle and the association is complete, the number of particles falls by the factor

n

, so

i = \frac{1}{n}

. With

i = 0.5

this gives

n = 2

, the benzoic acid molecules associating in pairs.

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Integer answer

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