QJEEM21Ch9b03: JEE Main 2021 Chemistry - Class 12 JEEM21Ch9b03 Solution: When 12.2 g of benzoic acid is dissolved in 100 g of water, the …

Question

Verified Solution • 2025–26 Session moderate

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 (K_f(H₂O) = 1.86 K kg mol^{$-$ 1}). The number (n) of benzoic acid molecules associated (assuming 100% association) is ___________.

Accepted Answer

\underset{\text{Benzoic acid}}{n \mathrm{PhCOOH}} \stackrel{\text { Association }}{\longrightarrow}(\mathrm{PhCOOH})_n

Assuming

100 \%

association (

\alpha=1

),

\Rightarrow i = 1 - α (1 - \frac{1}{n}) = \frac{1}{n} [∵ α + 1]

Now,

\Delta T_f=K_f \times m \times i

0 - (0.93) = 1.86 \times \frac{w_{B} \times 1000}{w_{A} \times M_{B}} \times \frac{1}{n}

\left[\because w_B=\right.

mass of

\mathrm{PhCOOH}=12.2 \mathrm{~g}

w_A=

mass of

\mathrm{H}_2 \mathrm{O}=100 \mathrm{~g}

M_B=

molar mass of

\left.\mathrm{PhCOOH}\right]

=122 \mathrm{~g} \mathrm{~mol}^{-1}

=1.86 \times \frac{12.2 \times 1000}{100 \times 122} \times \frac{1}{n}

\begin{aligned} \Rightarrow n & = \frac{1.86 \times 12.2 \times 1000}{0.93 \times 100 \times 122} = 2 \end{aligned}

\therefore

Number of benzoic acid molecules associated,

n=2

Step-by-Step Verified Solution