A solution of glucose (C6H12O6 ) is 18% by mass. The mole fraction of glucose will be

Molar mass of $C_{6} H_{12} O_{6}$ , $M_{2} = 180 g {mol}^{- 1}$

Mass of glucose in the 100 g solution m₂ = 18 g

Amount of glucose in 100 g solution $n_{2} = \frac{m_{2}}{M_{1}} = \frac{10 g}{180 g {mol}^{- 1}} = 0.1 mol$

Mass of water in 100 g solution, $m_{1} = 82 g$

Amount of water in 100 g solution $n_{1} = \frac{m_{1}}{M_{1}} = \frac{8 ∠ g}{18 g {mol}^{- 1}} = 4.56 mol$

Mole fraction of glucose, $x_{2} = \frac{n_{2}}{n_{1} + n_{2}} = \frac{0.1 mol}{(4.56 + 0.1) mol} = 0.0214$

A solution of glucose (C₆H₁₂O_{6 )} is 18% by mass. The mole fraction of glucose will be