Answer:
The molar mas of the X is 203.06 g/mol.
Explanation:
The pressure of the pure solvent = p = 23.8 Torr
Vapor pressure of the solution = [tex]p_s=23.5 Torr[/tex]
Mass of solute =14.4 g
Molar mass of solute = M
Moles of solute = [tex]n_1=\frac{14.4 g}{M}[/tex]
Mass of solvent or water = 100.0 g
Moles of water = [tex]n_2=\frac{100.0 g}{18 g/mol}=5.555 mol[/tex]
Mole fraction of solute = [tex]\chi_1=\frac{n_1}{n_1+n_2}[/tex]
The relative lowering in vapor pressure of the solution with non volatile solute is equal mole fraction of solute in solution.
[tex]\frac{p-p_s}{p}=\chi_2=\frac{n_1}{n_1+n_2}[/tex]
[tex]\frac{23.8 Torr-23.5Torr}{23.8Torr}=\frac{n_1}{n_1+5.555 mol}[/tex]
Solving for [tex]n_1[/tex]
[tex]n_1=0.070915 mol[/tex]
[tex]n_1=\frac{14.4 g}{M}[/tex]
[tex]M=\frac{14.4 g}{0.070915 mol}=20306 g/mol[/tex]
The molar mas of the X is 203.06 g/mol.
