Question
Asked in: JEE Main-2018
The mass of a non-volatile, non-electrolyte solute (molar mass=50 g ) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%, is :
A.
37.5 g
B.
75 g
C.
150 g
D.
12.5 g
Let massof non-volatile,non-electrolyte solute is w.
molar mass of solute , m = 50 g/mol
mass of solvent , W= mass of octane
=114g
molar mass of solvent , M = 114g/mol
first of all, find mole fraction of solute ;
e.g., mole fraction of solute = mole of solute/(mole of solute + mole of solvent )
= {w/50}/{w/50 + 114/114}
= w/(w + 50)
from lovering of vapor pressure,
?P/P = mole fraction of solute
According to the to question, vapor pressure is reduced by 75%
then, ?P/P = 0.75
now, 0.75 = w/(w + 50)
or, 3/4 = w/(w + 50)
or, 3w + 150 = 4w
or , w = 150g
hence, mass of non - volatile solute is 150g