Question

Vapour pressure of chloroform $ \left( {CHC{l_3}} \right) $ and dichloromethane $ \left( {C{H_2}C{l_2}} \right) $ at $ 298K $ are $ 200 $ mm Hg and $ 415 $ mm Hg respectively. Vapour pressure of the solution obtained by mixing $ 25.5g $ of $ CHC{l_3} $ and $ 40g $ of $ C{H_2}C{l_2} $ at the same temperature will be:
[Molecular mass of $ CHC{l_3} = 119.5g/mol $ and molecular mass of $ C{H_2}C{l_2} = 85g/mol $ ]
(A) $ 615 $ mm Hg
(B) $ 347.9 $ mm Hg
(C) $ 285.5 $ mm Hg
(D) $ 173.9 $ mm Hg

Answer 1

Hint: Vapour pressure of the solution is defined as the measure of the tendency of the material to change its state into a gaseous or vapor state. Simply, it is the pressure exerted by the vapors of the liquid over the liquid phase. We can easily calculate the vapor pressure of the mixture of the solution by considering their contribution.
Formula used
Raoult’s law, $ {p_T} = p_A^0{X_A} + p_B^0{X_B} $
Where $ p_A^0,p_B^0 $ are the vapor pressure of the solvents $ A $ and $ B $ respectively and $ {X_A},{X_B} $ are the mole fraction of $ A $ and $ B $ respectively.

Complete step by step answer:
First, we will understand the given quantities from the question. The question says that the vapor pressure of chloroform $ \left( {CHC{l_3}} \right) $ and dichloromethane $ \left( {C{H_2}C{l_2}} \right) $ $ 298K $ are $ p_A^0 = 200 $ mmHg and $ p_B^0 = 415 $ mm Hg respectively. We have also given that molar mass of $ CHC{l_3} = 119.5g/mol $ and molar mass of $ C{H_2}C{l_2} = 85g/mol $ . Now we will calculate the number of moles first. So the moles of $ 25.5g $ $ CHC{l_3} = \dfrac{{25.5}}{{119.5}} = 0.213mol $ and moles of $ 40g $ $ C{H_2}C{l_2} = \dfrac{{40}}{{85}} = 0.47mol $ . So, the total number of moles will be $ {n_A} + {n_B} = 0.47 + 0.213 = 0.683mol $ . Now, we will calculate the mole fraction of the compounds.
The mole fraction is given by $ {X_A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}},{X_B} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}} $ . So, the mole fraction of $ {X_{CHC{l_3}}} = \dfrac{{0.213}}{{0.683}} = 0.312,{X_{C{H_2}C{l_2}}} = \dfrac{{0.47}}{{0.683}} = 0.688 $ . Now we have the vapor pressure of the chloroform $ \left( {CHC{l_3}} \right) $ and dichloromethane $ \left( {C{H_2}C{l_2}} \right) $ and the mole fraction of the chloroform $ \left( {CHC{l_3}} \right) $ and dichloromethane $ \left( {C{H_2}C{l_2}} \right) $ . So, to calculate the total vapor pressure of the mixture using Raoult’s law, $ {p_T} = p_A^0{X_A} + p_B^0{X_B} $ . So, after substituting the values we get, $ {p_T} = p_A^0{X_A} + p_B^0{X_B} = 200 \times 0.312 + 415 \times 0.688 = 374.92 $ . So, the vapor pressure of the solution obtained by mixing $ 25.5g $ of $ CHC{l_3} $ and $ 40g $ of $ C{H_2}C{l_2} $ at the same temperature will be $ 374.92 $ mm Hg.
Therefore, the correct option is (B).

Note:
The vapor pressure of the solution is always less than the vapor pressure of the pure solvent. If the mole fraction of solute $ A $ is $ {X_A} $ then we can calculate the mole fraction of solute $ B $ which is $ {X_B} $ using the relation $ {X_A} + {X_B} = 1,{X_B} = 1 - {X_A} $ .