Question

Which of the following solutions will exhibit the highest boiling point?

• A. $0.01MKN{O}_3$
• B. $0.015M$ Glucose
• C. $0.015M$ Urea
• D. $0.01MN{a}_{2}S{O}_4$

Answer 1

The correct option is D $0.01mN{a}_{2}S{O}_4$

From boiling point of elevation,
$\Delta T=i\times K_b\times m$
where,
$△T_b$ is the elevation in boiling point.
$K_b$ is molal elevation constant.
m is molality of solution.
i is van't Hoff factor

From the equation we can say higher the value of $i\times m$ higher will be the boiling point elevation.
Considering all the compounds exhibit 100% dissociation,
$N{a}_{2}S{O}_4\to 2N{a}^{+}+S{O}_4^{2-}$
$KN{O}_3\to K^{+}+N{O}_3^{-}$
Urea and glucose are non-electrolyte solute
$$\begin{array}{cc}N{a}_{2}S{O}_4\Rightarrow & i\times m=3\times 0.01=0.03\\ KN{O}_3\Rightarrow & i\times m=2\times 0.01=0.02\\ \text{Urea}\Rightarrow & i\times m=1\times 0.015=0.015\\ \text{Glucose}\Rightarrow & i\times m=1\times 0.015=0.015\end{array}$$
Since, $i\times m$ is higher for $N{a}_{2}S{O}_4$, the boiling point will be higher for $0.01MN{a}_{2}S{O}_4$