Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point by 0.2$^\circ$C. The percentage association of solute A in water, is :
[Given : Molar mass of A = 93 g mol$-$1. Molal depression constant of water is 1.86 K kg mol$-$1.]
Solution
Since, $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik}_{\mathrm{f} m}$
<br/><br/>
$$
\begin{aligned}
&m=\frac{0.7}{93} \times \frac{1000}{42} \\\\
&0.2=i \times 1.86 \times \frac{0.7 \times 1000}{93 \times 42} \\\\
&i=0.6 \\\\
&\alpha=\frac{i-1}{\frac{1}{n}-1}=\frac{0.6-1}{\frac{1}{2}-1}=0.8
\end{aligned}
$$
<br/><br/>
Hence, the percentage association of solute $A$ is $80 \%$.
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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