JEE MAIN - Chemistry (2021 - 26th February Evening Shift - No. 17)
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(H2O) = 1.86 K kg mol$$-$$1). The number (n) of benzoic acid molecules associated (assuming 100% association) is ___________.
Answer
2
Explanation
$\underset{\text{Benzoic acid}}{n \mathrm{PhCOOH}} \stackrel{\text { Association }}{\longrightarrow}(\mathrm{PhCOOH})_n$
Assuming $100 \%$ association ( $\alpha=1$ ),
$$ \Rightarrow i=1-\alpha\left(1-\frac{1}{n}\right)=\frac{1}{n}[\because \alpha+1] $$
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, $n=2$
Assuming $100 \%$ association ( $\alpha=1$ ),
$$ \Rightarrow i=1-\alpha\left(1-\frac{1}{n}\right)=\frac{1}{n}[\because \alpha+1] $$
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, $n=2$
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