Q50 / 200 · Manipal (MET) 2012 · Chemistry

The elevation in boiling point would be highest for

  • A
    $0.08 \mathrm{M} \mathrm{BaCl}_2$
  • Correct answer
    B
    0.15 M KCl
  • C
    0.10 M Glucose
  • D
    $0.06 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_3\right)_2$

Solution

$\begin{array}{ll} \because & \Delta T_b=\frac{1000 \cdot K_b \cdot w}{m \cdot W} \times i \\ \therefore & \Delta T_b \propto i \cdot \frac{w}{m \cdot \frac{W}{1000}}\left(\because K_b=\text { constant }\right) \end{array}$ or $\quad \Delta T_b \propto i \cdot M$ $\begin{aligned} (\because \text { Molality } & =\frac{w}{m \cdot \frac{W}{1000}} \text { and assuming, } \\ & \text { Molarity, } M=\text { molality, } m) \end{aligned}$ Now, for the given solutions, (a) For $0.08 \mathrm{M} \mathrm{BaCl}_2, i \cdot M=3 \times 0.08=0.24$ (b) For $0.15 \mathrm{M} \mathrm{KCl}, i \cdot M=2 \times 0.15=0.30$ (c) For 0.10 M glucose, $i \cdot M=1 \times 0.10=0.10$ (d) For $0.06 \mathrm{MCa}\left(\mathrm{NO}_3\right)_2, i \cdot M=3 \times 0.06$ $=0.18$ Hence, $\Delta T_b$ is maximum for 0.15 M KCl solution.

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

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