Question

A solution of a non-volatile solute in water has a boiling point of $375.3K$. The vapour pressure of water above this solution at $338K$ is:
[Given ${p^o}(water) = 0.2467atm$ at $338K$ and ${K_b}$ for water $ = 0.52K\;kg\;mo{l^{ - 1}}$]
A. $0.18atm$
B. $0.23atm$
C. $0.34atm$
D. $0.42atm$

Answer 1

Hint:As we know that a non-volatile solute is a substance which does not readily evaporate into any gas under given conditions. It exhibits a low vapour pressure and high boiling point. Vapour pressure is the measure of change in the present state of a substance to its vapour state or gaseous state.

Formula used: $\Delta {T_b} = ({T^o} - {T^s})$, $mole\;fraction = \dfrac{{moles\;of\;solvent}}{{moles\;of\;solute + \;solvent}}$ and${P^s} = {p^0} \times mole\;fraction$

Complete answer:
We know that a non-volatile substance is the one which easily evaporates into gaseous form under given conditions of temperature and pressure. To calculate the vapour pressure we should have the mole fraction of the solute and solvent. Boiling point is reached when vapour pressure and atmospheric pressure becomes equal.
We are given with the boiling point of water using which we can first calculate the boiling point elevation temperature using the formula:
$\Delta {T_b} = ({T^o} - {T^s})$
$
  \Delta {T_b} = (375.3 - 373.16) \\
\Rightarrow \Delta {T_b} = 2.15K
 $
Now using this boiling point elevation temperature we can calculate the molality of the solute, we already have ${K_b}$ for water $ = 0.52K\;kg\;mo{l^{ - 1}}$, so using the formula:
\[\Delta {T_b} = {K_b} \times m\]
$\Rightarrow m = \dfrac{{\Delta {T_b}}}{{{K_b}}}$
$
\Rightarrow m = \dfrac{{2.15}}{{0.52}} \\
\Rightarrow m = 4.135
 $
Now using this molality we need to find out the mole fraction of the water to calculate the vapour pressure of water.
So we can say that $4.135$ moles of solute is present in $1\;kg$ or $55.5$ moles of solvent that is water. So the mole fraction will be:
$mole\;fraction = \dfrac{{moles\;of\;solvent}}{{moles\;of\;solute + \;solvent}}$
$\Rightarrow mole\,fraction = \dfrac{{55.5}}{{4.135 + 55.5}}$
$\Rightarrow mole\;fraction = \dfrac{{55.5}}{{59.635}}$
$\Rightarrow mole\;fraction = 0.931$
Now we know that vapour pressure of water can be calculate as the product of mole fraction and partial pressure and it can given as:
${P_s} = {p^{\circ}} \times mole\;fraction$
After putting all the given and calculated values we will get:
${P_s} = 0.2467 \times 0.931$
$\Rightarrow {P_s} = 0.23\;atm$

Therefore from the above explanation the correct answer is Option B.

Note:
1. Vapour pressure is the tendency of a substance to change into the vapour phase or gaseous state from its original state which increases with increase in water and the boiling point is that temperature at which the vapour pressure of the substance at the surface of a liquid becomes equivalent to the pressure exerted by the surroundings.
2. Remember that higher the vapour pressure of a substance, faster it will evaporate into vapours.