# QuestionAsked in: JEE Main-2018The 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 gB.75 gC.150 gD.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

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