Question

The mass of a non-volatile, non-electrolyte solute (molar mass=50 g mol1) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%, is :

Answer 1

Let mass of non - volatile , non - electrolyte solute is w.

molar mass of solute , m = 50 g/mol

mass of solvent , W= mass of octane (C8H18)

= 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

a/c to question, [here mistake happened, 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

Answer 2

Answer: its answer is given as 50..!?? how plzz explain...

Explanation: