Question

1 mole ${A_X}{B_Y}$ when added to $1.08kg$ of water, lowers its vapor pressure by $35torr$ to $700torr$ . If further $1184g$ of methanol is added, then the vapor pressure increases by $37torr$ . Considering only a few vapors are formed, and ${A_X}{B_Y}$ is a non-volatile solute. What is the value of $a$ if the vapor pressure of pure methanol is $a \times {10^2}torr$ .

Answer 1

Hint: The relative lowering in the vapor pressure is a colligative property which depends on the relative number of particles of solute and does not depend upon the nature of the particles. The colligative properties are directly proportional to the mole fraction of solute and thus, on the number of moles of solute.

Complete step by step answer:
According to the relative lowering of vapor pressure, it is equal to the mole fraction of the solute. Mathematically, it can be written as:
\[\dfrac{{{P^o} - P}}{{{P^o}}} = x\]….(i)
Where, \[{P^o} = \] Vapor pressure of pure solvent
$P = $ Vapor pressure of the solution
$x = $ mole fraction of the solute
Molar mass of water = ${H_2}O = 18$
Molar mass of methanol = $C{H_3}OH = 32$
Number of moles = $n = \dfrac{{{w_{given}}}}{{{M_w}}}$
Mole fraction = $x = \dfrac{n}{{n + N}}$
Where, $n = $ moles of solute
$N = $ moles of solvent
Number of moles of water = $\dfrac{{1.08 \times 1000}}{{18}} = 60moles$
Number of moles of methanol = $\dfrac{{1.184 \times 1000}}{{32}} = 37moles$
When only the solute ${A_X}{B_Y}$ is added to the water, the equation (i) becomes,
\[ \Rightarrow \dfrac{{{P^o} - P}}{{{P^o}}} = x\]
\[ \Rightarrow \dfrac{{735 - 700}}{{735}} = \dfrac{1}{{1 + 60}} = 0.0164\]…. (ii)
When methanol is also added in the solution, then equation (i) becomes,
$ \Rightarrow \dfrac{{{P^o} - 700}}{{{P^o}}} = \dfrac{1}{{1 + 60 + 37}} = 0.0102$
Thus, ${P^o} = 745torr$
But, this value of vapor pressure corresponds to only a mixture of methanol and water.
Let $P'$ be the partial vapor pressure of methanol in this mixture. From Dalton’s law of partial pressure, we have:
${P_{total}} = {P_A}{x_A} + {P_B}{x_B}$ … (iii)
Where, ${P_{total}} = $Total vapor pressure of the mixture = $745torr$
${P_A} = $ partial pressure of water = $735torr$
${P_B} = $ partial pressure of methanol = $P'$
${x_A} = $mole fraction of water = $\dfrac{{60}}{{37 + 60}} = \dfrac{{60}}{{97}}$
${x_B} = $mole fraction of methanol = $\dfrac{{37}}{{37 + 60}} = \dfrac{{37}}{{97}}$
Substituting these values in (iii), we have:
$ \Rightarrow 745 = 735 \times \dfrac{{60}}{{97}} + P' \times \dfrac{{37}}{{97}}$
$ \Rightarrow P' = 761torr \approx 8 \times {10^2}torr$
Comparing this value with the given value of $a \times {10^2}$ , we have the value of $a = 8$ .

Note:
When a non-volatile solute is dissolved in a pure solvent, the vapor pressure of the solvent is lowered i.e. the vapor pressure of a solution is always lower than the vapor pressure of a pure solvent, because the escaping tendency of solvent molecules decreases (due to lesser solvent molecules per unit surface area).