Question

Calculate the molal elevation constant, ${k_b}$ for water and the boiling point of 0.1 molal urea solution. It is given that latent heat of vaporization of water is $9.72kcal/mol$ at $373.15K$.
A) ${K_b} = 0.515Kkg/mol,{T_b} = 373.20K$
B) ${K_b} = 0.515Kkg/mol,{T_b} = 370.20K$
C) ${K_b} = 0.565Kkg/mol,{T_b} = 373.20K$
D) ${K_b} = 5.15Kkg/mol,{T_b} = 370.20K$

Answer 1

Find the molal elevation constant by using its formula by putting the values boiling point of pure solvent and latent heat of vaporization. Then, find the depression in the freezing point and add with the given temperature in the question.

Complete step by step solution:
The formula of molal elevation constant is –
${K_b} = \dfrac{{0.002 \times {{\left( {T_b^o} \right)}^2}}}{{{L_v}}}$
where, $T_b^o$ is the boiling point of pure solvent and ${L_v}$ is the latent heat of vaporization.
Complete Step by Step Solution: -
First of all, we have to find the molal elevation constant, so, we know that, molal elevation constant is given by –
${K_b} = \dfrac{{0.002 \times {{\left( {T_b^o} \right)}^2}}}{{{L_v}}} \cdots \left( 1 \right)$
where, $T_b^o$ is the boiling point of pure solvent and ${L_v}$ is the latent heat of vaporization in $cal/g$ of the pure solvent.
Therefore, from the question, it is given that –
Latent heat of vaporization, ${L_v} = 9.72kcal/mol = 540cal/g$
So, in equation (1), putting the value of latent heat vaporization as $540cal/g$ and boiling point of pure solvent as 373.15, we get –
$\Rightarrow {K_b} = \dfrac{{0.002 \times {{\left( {373.15} \right)}^2}}}{{540}} \\\ \Rightarrow {K_b} = 0.515Kkg/mol \\\$
Now, calculating the depression in freezing point which is given by the formula –
$\Delta {T_b} = {K_b} \times m$
Putting the value of molal elevation constant and molal urea solution in the above formula, we get –
$\\\Rightarrow \Delta {T_b} = 0.515 \times 0.1 \\\ \therefore \Delta {T_b} = 0.0515 \\\$
So, now the boiling point can be calculated by using the value of depression in freezing point and given temperature in the question –
$\Rightarrow {T_b} = 373.15 + 0.0515 \approx 373.20K$
Hence, we got the molal elevation constant as $0.515Kkg/mol$ and boiling point as $373.20K$.
Hence, the correct option is (A) .

Note: The boiling purpose of a substance is that the temperature at that the vapour pressure of the liquid equals the pressure surrounding the liquid and therefore the liquid changes into a vapor. The boiling point of a liquid varies relying upon the encircling environmental pressure.