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 C (Kf( H_2O ) = 1.86 K kg mol - 1). The number (n) of benzoic acid molecules associated (assuming 100% association) is \underline\hspace2cm .

Solutions PYQ — JEE Main 2021 · 26 Feb · Shift 2 · Q17 with Solution

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Correct answer: 2

For a solute in solution,

\Delta T_f = i K_f m

where:

Since pure water freezes at $0^\circ\text{C}$ and solution freezes at $-0.93^\circ\text{C}$ ,

\Delta T_f = 0.93

Molar mass of benzoic acid $\left(\mathrm{C_7H_6O_2}\right)$ :

7\times 12 + 6\times 1 + 2\times 16 = 84 + 6 + 32 = 122\,\text{g mol}^{-1}

Moles of benzoic acid:

\frac{12.2}{122} = 0.1\,\text{mol}

Mass of water:

100\,\text{g} = 0.1\,\text{kg}

So molality is:

m = \frac{0.1}{0.1} = 1\,\text{mol kg}^{-1}

Using

\Delta T_f = iK_fm

0.93 = i \times 1.86 \times 1

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

If $n$ molecules associate completely to form one aggregate, then starting from $n$ molecules, number of particles becomes $1$ .

Hence for 100% association,

i = \frac{1}{n}

Given $i=0.5$ ,

\frac{1}{n} = 0.5

n = 2

The benzoic acid molecules associate as a dimer.

\boxed{2}