Q1 / 200 · Manipal (MET) 2013 · Chemistry

0.01 M solution of KCl and $\mathrm{CaCl}_2$ are prepared in water. The freezing point of KCl is found to be $-2^{\circ} \mathrm{C}$. What is the freezing point of $\mathrm{CaCl}_2$ to be completely ionised?

  • Correct answer
    A
    $-3^{\circ} \mathrm{C}$
  • B
    $+3^{\circ} \mathrm{C}$
  • C
    $-2^{\circ} \mathrm{C}$
  • D
    $-4^{\circ} \mathrm{C}$

Solution

$i$ for $\mathrm{KCl}=2, i$ for for $\mathrm{CaCl}_2=3$ $\begin{aligned} & \frac{\Delta T_f(\mathrm{KCl})}{\Delta T_f \propto i}=\frac{2}{\Delta T_f\left(\mathrm{CaCl}_2\right)}=\frac{2}{3}\end{aligned}$ $\Delta T_f\left(\mathrm{CaCl}_2\right)=\frac{3}{2} \times 2=3^{\circ} \mathrm{C}$ Freezing point of $\mathrm{CaCl}_2=-3^{\circ} \mathrm{C}$

Hence, the correct answer is option: 'A'.

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