Question

The elevation in boiling point would be highest for

• A. $0.08\, M\, BaCl_{2}$
• B. $0.15\, M\, KCl$
• C. $0.10\,M$ Glucose
• D. $0.06\, M\, Ca(NO_{3})_{2}$

Answer 1

Answer: (B)

$\because \Delta T_{b}=\frac{1000 \cdot K_{b} \cdot w}{m \cdot W} \times i$
$\therefore \Delta T_{b} \propto i \cdot \frac{w}{m \cdot \frac{W}{1000}}$
$\left(\because K_{b}=\right.$ constant $)$
or $\Delta T_{b} \propto i \cdot M$
$\left(\because\right.$ Molality $=\frac{w}{m \cdot \frac{W}{1000}}$ and assuming,
Molarity, $M=$ molality, $m)$
Now, for the given solutions,
(a) For $0.08\, M\, BaCl _{2}, i \cdot M=3 \times 0.08=0.24$
(b) For $0.15\, M\, KCl , i \cdot M=2 \times 0.15=0.30$
(c) For $0.10\, M$ glucose, $i \cdot M=1 \times 0.10=0.10$
(d) For $0.06\, M\, Ca \left( NO _{3}\right)_{2}, i \cdot M=3 \times 0.06$
$=0.18$
Hence, $\Delta T_{b}$ is maximum for $0.15\, M\, KCl$ solution.