Question

The molal elevation constant of water is $0.51$. The boiling point of $0.1$ molal aqueous ${\text{NaCl}}$ solution is nearly:
(A) $100.05{\text{ }}^\circ {\text{C}}$
(B) $100.1{\text{ }}^\circ {\text{C}}$
(C) $100.2{\text{ }}^\circ {\text{C}}$
(D) $101.0{\text{ }}^\circ {\text{C}}$

Answer 1

When any non volatile solute is mixed in a solvent, it increases the boiling point of solvent. Using the given molal elevation constant of water and the molality of solution, we can find the increase in boiling point.

Formula Used:
$\Delta {T_b} = i \cdot {K_b} \cdot m$

Complete step by step answer:
It is given that:
Molal elevation constant of water, ${K_b} = 0.51{\text{ K kg mo}}{{\text{l}}^{ - 1}}$
Molality of aqueous solution, $m = 0.1{\text{ mol k}}{{\text{g}}^{ - 1}}$
${\text{NaCl}}$is a strong electrolyte and it dissociates in water as:
${\text{NaCl}} \to {\text{N}}{{\text{a}}^ + } + {\text{C}}{{\text{l}}^ - }$
One molecule of ${\text{NaCl}}$ produces two particles in the solution.
Thus, van’t Hoff factor, $i = 2$
Increase in boiling point, $\Delta {T_b} = ?$
Relationship between above four quantities is:
$\Delta {T_b} = i \cdot {K_b} \cdot m$
Putting the values of ${K_b}$, $m$ and $i$ in this expression, we get:
$\Delta {T_b} = 2 \times 0.51 \times 0.1{\text{ K}}$
Solving the above expression:
$\Delta {T_b} = 0.102{\text{ K}}$
This numerical value of difference of temperature will remain the same in any temperature scale. Thus, we can also write, $\Delta {T_b} = 0.102{\text{ }}^\circ {\text{C}}$
Now, at standard conditions, the boiling point of pure water, $T_b^0 = 100{\text{ }}^\circ {\text{C}}$
Boiling point of solution, ${T_b} = T_b^0 + \Delta {T_b}$
Substituting the values of $T_b^0$ and $\Delta {T_b}$, we get:
${T_b} = (100 + 0.102){\text{ }}^\circ {\text{C}}$
$\Rightarrow {T_b} = 100.102{\text{ }}^\circ {\text{C}}$
Rounding off the above value upto one decimal place, we get:
${T_b} = 100.1{\text{ }}^\circ {\text{C}}$

Hence, the correct option is (B).

Note: In case of non-electrolyte solutions, we can take the value of van’t Hoff factor, $i = 1$. But in this example, we cannot ignore the value of $i$.
We have calculated the elevation in boiling point (difference of temperature) as ${\text{0}}{\text{.102 K}}$. The numerical value of this difference of temperature will remain the same in any temperature scale. For instance, if we want to calculate in centigrade scale:$\Delta {T_b} = 100.102{\text{ }}^\circ {\text{C}} - 100{\text{ }}^\circ {\text{C}} = 0.102{\text{ }}^\circ {\text{C}}$