Question

If ${P^0}$ and P are the vapor pressure of the pure solvent and solution and ${n_1}$ and ${n_2}$ are the moles of solute and solvent respectively in the solution then the correct relation between P and ${P^0}$ :
 A.${P^0} = P[\dfrac{{{n_1}}}{{{n_1} + {n_2}}}]$
B. ${P^0} = P[\dfrac{{{n_2}}}{{{n_1} + {n_2}}}]$
C.$P = {P^0}[\dfrac{{{n_2}}}{{{n_1} + {n_2}}}]$
D. $P = {P^0}[\dfrac{{{n_1}}}{{{n_1} + {n_2}}}]$

Answer 1

Hint: When a solute is added to a solvent, the vapor pressure of solvent is lower than the vapor pressure above pure solvent. If the vapor pressure is low in the colligative property of solutions and the vapor pressure of a pure solvent is greater than the vapor pressure of a solution containing a nonvolatile liquid.

Complete step by step answer:
The solute is added in solvent and makes a solution. The vapor pressure of solvent is greater than the vapor pressure of solution.
This is followed by Raoult's law.
$
  \dfrac{{{P^0} - P}}{{{P^0}}} = \dfrac{{{n_1}}}{{{n_1} + {n_2}}} \\
  1 - \dfrac{P}{{{P^0}}} = \dfrac{{{n_1}}}{{{n_1} + {n_2}}} \\
  \dfrac{P}{{{P^0}}} = 1 - \dfrac{{{n_1}}}{{{n_1} + {n_2}}} \\
  \dfrac{P}{{{P^0}}} = \dfrac{{{n_2}}}{{{n_1} + {n_2}}} \\
  P = {P^0}[\dfrac{{{n_2}}}{{{n_1} + {n_2}}}] \\
 $

Hence, option (C) is the correct answer.

Additional information:
When the vapor pressure of the solvent above a solution containing nonvolatile solute is directly proportional to the mole fraction of solvent in the solution. A volatile solute will contribute to the vapor pressure above a solution in which it dissolves. The vapor pressure above a solution containing volatile solute is equal to the sum of vapor pressures of the solvent and volatile solutes.

Note:
Raoult’s law applies to solutions in which solute is nonvolatile and hasn’t got a tendency to form a vapor at the temperature of the solution, it goes on to explain how the result at low of vapor pressure affects the boiling point and freezing point of the solution.