The mole fraction of glucose in 0.15 M solution (density= 1.10 g mL^-1 ) is

Solution:

Let we have

Amount of glucose, n₂ = 0.15 mol;   Volume of solution, V = 1000 mL

Mass of glucose, m₂ = n₂M₂ = (0.15 mol) (180 g mol ^-1) = 27.0 g.

Mass of solution, m = V$\rho$ = (1000 mL) (1.10 g mL^-1 ) = 1100 g

Mass of water, m₁ = m₁ - m₂ = 1100 g - 27 g = 1073 g

Amount of water, $n_1=\frac{m_1}{M_1}=\frac{10/3\mathrm{g}}{18\mathrm{g}{\mathrm{mol}}^{-1}}=59.61\mathrm{mol}$

Mole fraction of glucose, $x_2=\frac{n_2}{n_1+n_2}=\frac{0.15\mathrm{mol}}{(59.61+0.15)\mathrm{mol}}=0.002$