The mass of ethanol (molar mass $=46\mathrm{g}{\mathrm{mol}}^{-1}$) to be added to $1.0\mathrm{kg}$ of water so as to have its amount fraction equal to $0.2$ is

By definition $x_2=n_2/\left(n_1+n_2\right)$

Hence, $x_2\left(n_1+n_2\right)=n_2\text{ or }{x}_2{n}_1=\left(1-x_2\right)n_2\text{ or }{x}_2{n}_1=x_1{n}_2\text{ or }{n}_2=x_2{n}_1/x_1$

Now, $n_1=\frac{m_1}{M_1}=\frac{1000\mathrm{g}}{18\mathrm{g}{\mathrm{mol}}^{-1}}=55.56\mathrm{mol}$, Hence ${\mathrm{n}}_2=(0.2)(55.56\mathrm{mol})/(0.8)=13.89\mathrm{mol}$

Mass of ethanol required is ${\mathrm{m}}_2={\mathrm{n}}_2{\mathrm{M}}_2=(13.89\mathrm{mol})\left(46\mathrm{g}{\mathrm{mol}}^{-1}\right)=638.9\mathrm{g}$