Answer
$$x_{glycerol} = 0.03593$$
Work Step by Step
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 = $$84.00 \space g \times \frac{1 \space mol}{18.02 \space g} = 4.661
\space mol$$
$$x_{glycerol} = \frac{0.1737 \space mol}{0.1737 \space mol + 4.661 \space mol} = 0.03593$$
