Question

Vapour pressure of chloroform ${\text{(CHC}}{{\text{l}}_{\text{3}}}{\text{)}}$ and dichloromethane $({\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}{\text{)}}$ at {\text{25^\circ C}}are ${\text{200 mm Hg}}$ and ${\text{41}}{\text{.5 mm Hg}}$respectively. Vapour pressure of the solution obtained by mixing $25.5{\text{ g}}$ of ${\text{CHC}}{{\text{l}}_{\text{3}}}$and ${\text{40 g}}$of ${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$at the same temperature will be:
[Molecular mass of ${\text{CHC}}{{\text{l}}_{\text{3}}} = {\text{119}}{\text{.5 u}}$and molecular mass of${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}} = 85{\text{ u}}$]

A.615.0 mm Hg
B.347.9 mm Hg
C.285.5 mm Hg
D.173.9 mm Hg
E.90.38 mm Hg

Answer 1

The Dalton’s law states that when two or more gases which do not react chemically are kept in a closed space, the total pressure exerted by the mixture is equal to the sum of the partial pressure of individual gases. Use Dalton’s equation related to mole fraction, partial pressure and total pressure of the gas.

Step by step answer: We have given weights of individual gases so calculate the moles of each gas as follows:

${\text{moles (}}n{\text{) = }}\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$
To calculate the moles of chloroform ${\text{(CHC}}{{\text{l}}_{\text{3}}}{\text{)}}$substitute $25.5{\text{ g}}$ for mass and ${\text{119}}{\text{.5 u}}$ for molar mass.
${n_{{\text{CHC}}{{\text{l}}_{\text{3}}}}}{\text{ = }}\dfrac{{{\text{25}}{\text{.5 g}}}}{{119.5{\text{ u}}}}$
${n_{{\text{CHC}}{{\text{l}}_{\text{3}}}}} = {\text{0}}{\text{.213 mol}}$

Similarly, calculate the moles of dichloromethane

$({\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}{\text{)}}$ by substituting ${\text{40 g}}$ for mass and ${\text{85 u}}$ for molar mass.
${n_{{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}}}{\text{ = }}\dfrac{{{\text{40 g}}}}{{85{\text{ u}}}}$
${{\text{n}}_{{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}}} = {\text{0}}{\text{.47 mol}}$

To calculate the total pressure we need mole fractions of each gas. So, calculate the mole fraction of
Chloroform ${\text{(CHC}}{{\text{l}}_{\text{3}}}{\text{)}}$and dichloromethane $({\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}{\text{)}}$ as follows:
${\text{mole fraction }}({X_A}){\text{ = }}\dfrac{{{n_A}}}{n}$
Where,
${X_A}$= mole fraction of gas A
${n_A}$= moles of gas A
$n$= total moles of mixture

$n = {n_{{\text{CHC}}{{\text{l}}_{\text{3}}}}} + {n_{{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}}} = {\text{0}}{\text{.213 mol + 0}}{\text{.47 mol}} = {\text{0}}{\text{.683 mol}}$

So, to calculate the mole fraction of chloroform $({X_{{\text{CHC}}{{\text{l}}_{\text{3}}}}})$ substitute 0.213 mol for a mole of chloroform and 0.683 mol for total moles of the mixture.
${X_{{\text{CHC}}{{\text{l}}_{\text{3}}}}} = \dfrac{{{n_{{\text{CHC}}{{\text{l}}_{\text{3}}}}}}}{n} = \dfrac{{{\text{0}}{\text{.213 mol}}}}{{{\text{0}}{\text{.683 mol}}}} = 0.31$

Similarly, to calculate the mole fraction of dichloromethane

$({X_{{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}}})$ substitute 0.47 mol for a mole of dichloromethane and 0.683 mol for total moles of the mixture.
${X_{{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}}} = \dfrac{{{n_{_{{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}}}}}}{n} = \dfrac{{{\text{0}}{\text{.47 mol}}}}{{{\text{0}}{\text{.683 mol}}}} = 0.69$
Now, using mole fractions and partial pressure of each gas calculate the total vapour pressure of mixture as follows:

$P = {X_{{\text{CHC}}{{\text{l}}_{\text{3}}}}}{P_{{\text{CHC}}{{\text{l}}_{\text{3}}}}} + {X_{{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}}}{P_{{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}}}$
$P = (0.31 \times {\text{200 mm Hg}}) + (0.69 \times {\text{41}}{\text{.5 mm Hg}})$
$P = {\text{ 90}}{\text{.63 mm Hg}}$

So, the total vapour pressure of the solution is 90.63 mm Hg.

As option E is close to the answer so the correct option is (E) 90.38 mm Hg.

Note: The sum of the mole fraction is always 1. So in a mixture of two gases after calculating mole fraction of one gas we can calculate the mole fraction of another gas by subtracting calculated moles of gas from 1.