Question

The elevation in boiling point would be highest for

• A. $0.08MBaC{l}_2$
• B. $0.15MKCl$
• C. $0.10M$ Glucose
• D. $0.06MCa(N{O}_3{)}_2$

Answer 1

Answer: (B)

$\because \Delta T_b=\frac{1000\cdot K_b\cdot w}{m\cdot W}\times i$
$\therefore \Delta T_b\propto i\cdot \frac{w}{m\cdot \frac{W}{1000}}$
$(\because K_b=$ constant $)$
or $\Delta T_b\propto i\cdot M$
$(\because$ Molality $=\frac{w}{m\cdot \frac{W}{1000}}$ and assuming,
Molarity, $M=$ molality, $m)$
Now, for the given solutions,
(a) For $0.08MBaC{l}_2,i\cdot M=3\times 0.08=0.24$
(b) For $0.15MKCl,i\cdot M=2\times 0.15=0.30$
(c) For $0.10M$ glucose, $i\cdot M=1\times 0.10=0.10$
(d) For $0.06MCa{\left(N{O}_3\right)}_2,i\cdot M=3\times 0.06$
$=0.18$
Hence, $\Delta T_b$ is maximum for $0.15MKCl$ solution.