Question

Calculate the mass of non volatile solute which should be dissolved in hundred and 14 gram of h2o octane to reduce its vapour pressure to 80%

Answer 1

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

molar mass of solute , m = 40 g/mol [you forgot to type molar mass of solute i.e., 40g/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/40}/{w/40 + 114/114}

= w/(w + 40)

from lovering of vapor pressure,

∆P/P = mole fraction of solute

a/c to question, vapor pressure reduced to 80%

so, ∆P/P = (P - 0.80P)/P = 0.20

now, 0.2 = w/(w + 40)

or, 1/5 = w/(w + 40)

or, w + 40 = 5w

or, 4w = 40

or, w = 10g

hence, mass of solute is 10g