The mole fraction of methanol (CH3OH) in aqueous solution is 0.10. If density of the solution is 0.97 g mL-1 , the molarity of the solution will be

For x₂ = 0.10, let we have n₂ = 0.1 mol and n₁= 0.9 mol

The mass of solution will be

$m = n_{1} M_{1} + n_{2} M_{2} = (0.9 mol) (18 g {mol}^{- 1}) + (0.1 mol) (32 {mol}^{- 1}) = 19.4 g$

The volume of solution will be

$V = \frac{m}{ρ} = \frac{19.4 g}{0.97 {gmL}^{- 1}} = 20.0 ml = 20 \times 10^{- 3} L$

The molarity of solution will be

$M = \frac{n_{2}}{V} = \frac{0.1 mol}{20 \times 10^{- 3} L} = 5.0 {molL}^{- 1}$

The mole fraction of methanol (CH₃OH) in aqueous solution is 0.10. If density of the solution is 0.97 g mL^-1 , the molarity of the solution will be