Question

A sugar syrup (containing sugar and water only) of weight 214.2 g contains 34.2 g of sugar ($C_{12}{H}_{22}{O}_{11}$). Find the mole fraction of the sugar in the syrup.

• A. $0.1\times {10}^{-2}$
• B. $9.9\times {10}^{-3}$
• C. $9.9\times {10}^{-4}$
• D. $0.1\times {10}^{-3}$

Answer 1

The correct option is B $9.9\times {10}^{-3}$

Mass of sugar in the syrup = 34.2 g
Mass of water in syrup = 214.2 - 34.2 = 180 g
Molar mass of sugar ($C_{12}{H}_{22}{O}_{11}$) = 342 g
Number of moles = $\frac{\text{given mass}}{\text{molar mass}}$
Moles of sugar = $n_{sugar}=\frac{34.2}{342}$ = 0.1 mol
Moles of water = $n_{water}=\frac{180}{18}$ = 10 mol
Mole fraction = $\frac{\text{number of moles}}{\text{total number of moles}}$
$x_{sugar}=\frac{n_{sugar}}{n_{sugar}+n_{water}}$ = $\frac{0.1}{10.1}=9.9\times {10}^{-3}$