Question

Asked in: JEE Main-2018

The mass of a non-volatile, non-electrolyte solute (molar mass=50 g mol^{-1}) 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

Answers (1)

Let massof non-volatile,non-electrolyte solute is w.

molar mass of solute , m = 50 g/mol 

mass of solvent , W= mass of octane C_8H_{18}

=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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