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

Hint: 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:\[\Delta {T_b} = 100.102{\text{ }}^\circ {\text{C}} - 100{\text{ }}^\circ {\text{C}} = 0.102{\text{ }}^\circ {\text{C}}\]