$n_{H_{2}}=\frac{1000}{18}=55.5\, \mathrm{mol} \,H_{2} O$
$\mathrm{X}_{\text {solute }}=\frac{\mathrm{n}_{\text {calle }}}{\mathrm{n}_{\text {solute }}+\mathrm{n}_{\mathrm{H}_{2} \mathrm{O}}}=\frac{1}{1+55 \cdot 5}=0. 0177$
(ઉપયોગ $R$ $=0.083\,L\,bar\,K ^{-1}\,mol ^{-1}$ )