Question

The boiling point of an aqueous solution of a non-volatile solute is ${100.15^o}C$. What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal volume of water? /the value of ${K_b}$ and ${K_f}$ are $0.512$ and 1.86K$$$$molalit{y^{ - 1}}.
A.$- {0.544^o}C$
B.$- {0.512^o}C$
C.$- {0.272^o}C$
D.$- {1.86^o}C$

Answer 1

We need to know that the boiling of a compound is the temperature at which the pressure present in the surrounding is equal to the vapour pressure of a liquid and the phase of that compound changes from liquid state into vapour state. While boiling a substance, the particle present in the compound gains more energy and there is an increase in the kinetic energy of the particle. Hence, it can move very fast. And in the case of a non –
volatile substance, it will not readily evaporate into gas.

Complete answer:
We have to know that the freezing point of solution is not equal to $- {0.544^o}C$. Hence, option (A) is incorrect.
By substituting the given values in the equation of freezing point, it will not get as $- {0.512^o}C$. Hence, option (B) is incorrect.
According to the given question,
The boiling point of the non – volatile solute is equal to, {T_b}$$$${100.15^o}C. Hence,
$\Delta {T_b} = 100.15 - 100$
$= {0.15^o}C$
${K_b}$ and ${K_f}$ are $0.512$ and 1.86K$$$$molalit{y^{ - 1}}. Where, ${K_b}$is the elevation of boiling point and ${K_f}$is the depression in freezing point.
And the change on boiling point, $\Delta {T_b}$can be find out by using the equation,
$\Delta {T_b} = {K_b} \times M$
Where, M is molarity. Substitute the values in above equation, will get
$0.15 = 0.15 \times M$
Hence, $M = \dfrac{{0.512}}{{0.15}}$and
${M^{'}} = \dfrac{1}{2}\left( {\dfrac{{0.512}}{{0.15}}} \right)$
The $\Delta {T_f}$ can be find out by, $\Delta {T_f} = {K_f} \times {M^{'}}$
Substitute the values in above equation,
$\Delta {T_f} = 1.86 \times \dfrac{1}{2}\left( {\dfrac{{0.512}}{{0.15}}} \right)$
On simplification we get,
$= {0.272^o}C$
The freezing point of the solution can be found by subtracting the value of $\Delta {T_f}$ from the freezing point of water, which is equal to ${0^o}C$. Hence,
${T_f} = F.Pofwater - \Delta {{\rm T}_f}$
Now we can substitute the known values we get,
$= 0 - {0.272^o}C$
$= - {0.272^o}C$
Therefore, the freezing point of the aqueous solution is equal to $- {0.272^o}C$. Hence, option (C) is correct.
The freezing point of aqueous solution is not equal to $- {1.86^o}C$. Hence, option (D) is incorrect.

Hence, option (C) is correct.

Note:
The non – volatile solute will not vaporize or evaporate readily. And it always exhibits a high boiling point and low vapour pressure. Salt, sugar, etc are the examples of non – volatile solutes. In the case of volatile solutes, at the boiling point of the solution, it will produce the vapours. And it has higher vapour pressure than the volatile solutes at the same temperature. Alcohol, mercury, ether, etc are the examples of volatile solutes.