Question

A $2.0\%$ solution by weight of urea in water shows a boiling point elevation 0.18 deg [molecular weight of urea = 60], Calculate the latent heat of vaporization per gram of water.
A.495.3 cal/g
B.560.3 cal/g
C.525.8 cal/g
D.501.3 cal/g

Answer 1

Latent heat of vaporization is related to boiling point elevation because of the term molal elevation. We can know the value of molal elevation from molality and elevation in boiling point. And using that value, we can calculate latent heat of vaporization per gram of water.

Complete step by step answer:
The latent heat of vaporization is the amount of heat required to change 1 mole of liquid to vapour at its boiling point under STP conditions. We know that pure water boils at 100 deg at 1 atm pressure but the boiling point increases if we add a small amount of salt or sugar to it.
Adding any form of non-volatile solute to the liquid increases its boiling point and we call it elevation in boiling point. The amount by which the boiling point elevates depends directly on the amount of solute added.
Let us suppose $T_b^0$ is the boiling point temperature of pure liquid and ${T_b}$ is the boiling point of mixture, then we can state that $\Delta {T_b} = {K_b} \times m$ where ${K_b}$ is the molal elevation or ebullioscopy constant that depends on solvent, m is the molality and $\Delta {T_b} = T_b^0 - {T_b}$ i.e. elevation in boiling point.
From the problem, we see that $2\%$ solution by weight of urea means that 2 grams of urea is present in 100ml solution, so the mass of the solvent now becomes 98 grams. Molality can be calculated by the below formula:
$molality = \dfrac{{given\,\,mass/molar\,\,mass}}{{mass\,\,of\,\,solvent(in\,\,kg)}}$
Putting the values in it, we get
$molality = \dfrac{{2/60}}{{0.098}} = 0.34m$
Now calculating ${K_b}$ from $\Delta {T_b} = {K_b} \times m$ as we are provided $\Delta {T_b} = 0.18$
$0.18 = {K_b} \times 0.34$ or ${K_b} = 0.529$
Latent heat of vaporization is related to elevation in boiling point by the following formula:
${K_b} = \dfrac{{RT{{_b^0}^{^2}}}}{{1000{L_v}}}$ where R is the universal gas constant and ${L_v}$ is the latent heat of vaporization.
Now ${L_v} = \dfrac{{RT{{_b^0}^{^2}}}}{{1000{K_b}}}$ , putting the values in it, we get
${L_v} = \dfrac{{2 \times {{(373.15)}^2}}}{{1000 \times 0.529}}$
$\therefore$ ${L_v} = 525.8\,\,cal/g$

Hence, the correct option is (C).

Note:
There is one basic formula of latent heat of vaporisation in case we are given the mass of substance in kilograms and energy in joules, then latent heat of vaporisation in $J/Kg$ can be calculated by using this formula: $E = m \times {L_v}$.