Answer
(a)$$x_{CH_3CH_2OH} = 0.0977$$
(b) $$x_{urea} = 0.0122 $$
Work Step by Step
(a) 100 g of solution: 21.7 g of $CH_3CH_2OH$ and 78.3 g of $H_2O$.
$ CH_3CH_2OH $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 46.07 g/mol
$$ 21.7 \space g \times \frac{1 \space mole}{ 46.07 \space g} = 0.471 \space mole$$
$ H_2O $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol
$$ 78.3 \space g \times \frac{1 \space mole}{ 18.02 \space g} = 4.35 \space moles$$
$$x_{CH_3CH_2OH} = \frac{0.471 }{4.35 + 0.471} = 0.0977$$
(b) 0.684 mol of urea = 1 kg of water = 1000 g of water
$$1000 \space g \space H_2O \times \frac{1 \space mol}{18.02 \space g} = 55.49 \space mol $$
$$x_{urea} = \frac{0.684}{0.684 + 55.49} = 0.0122 $$
