JEE MAIN - Chemistry (2022 - 25th June Evening Shift - No. 3)
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.]
50%
60%
70%
80%
Explanation
Since, $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik}_{\mathrm{f} m}$
$$ \begin{aligned} &m=\frac{0.7}{93} \times \frac{1000}{42} \\\\ &0.2=i \times 1.86 \times \frac{0.7 \times 1000}{93 \times 42} \\\\ &i=0.6 \\\\ &\alpha=\frac{i-1}{\frac{1}{n}-1}=\frac{0.6-1}{\frac{1}{2}-1}=0.8 \end{aligned} $$
Hence, the percentage association of solute $A$ is $80 \%$.
$$ \begin{aligned} &m=\frac{0.7}{93} \times \frac{1000}{42} \\\\ &0.2=i \times 1.86 \times \frac{0.7 \times 1000}{93 \times 42} \\\\ &i=0.6 \\\\ &\alpha=\frac{i-1}{\frac{1}{n}-1}=\frac{0.6-1}{\frac{1}{2}-1}=0.8 \end{aligned} $$
Hence, the percentage association of solute $A$ is $80 \%$.
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