## General Chemistry: Principles and Modern Applications (10th Edition)

$$x_{glycerol} = 0.03593$$
By definition: 16.00% aqueous solution of glycerol by mass. 16.00 g of glycerol = 100 g of solution mass of water = 100 - 16.00 = 84.00 g of water. molar mass of glycerol: $C_3O_3H_8$ : ( 1.008 $\times$ 8 )+ ( 12.01 $\times$ 3 )+ ( 16.00 $\times$ 3 )= 92.09 g/mol moles of glycerol = $$16.00 \space g \times \frac{1 \space mol}{92.09 \space g} = 0.1737 \space mol$$ molar mass of water: $H_2O$ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol moles of water = $$84.00 \space g \times \frac{1 \space mol}{18.02 \space g} = 4.661 \space mol$$ $$x_{glycerol} = \frac{0.1737 \space mol}{0.1737 \space mol + 4.661 \space mol} = 0.03593$$