250 mL of a solution containing 10.0 g of sodium chloride and glucose produces 15 atm of osmotic pressure at 27 °C. The mass percentage of sodium chloride in the mixture is

Let m be the mass of $\mathrm{NaCl}$ in the mixture. 

Amount of $\mathrm{NaCl},n_{\mathrm{NaCl}}=\frac{m_{\mathrm{NaCl}}}{M_{\mathrm{NaCl}}}=\frac{m}{58.5\mathrm{g}{\mathrm{mol}}^{-1}}$ Amount of glucose, $n_{\text{glucose }}=\frac{10.0\mathrm{g}-m}{180\mathrm{g}{\mathrm{mol}}^{-1}}$

Total amount of species in the mixture

$n_{\text{total }}=\left\{n_{{\mathrm{Na}}^{+}}+n_{{\mathrm{Cl}}^{-}}\right\}+n_{\text{glucose }}=\left(\frac{2m}{58.5}+\frac{10.0\mathrm{g}-m}{180}\right)\frac{1}{\mathrm{g}{\mathrm{mol}}^{-1}}$Total concentration of species m the solution, $c=\frac{\left\{\right(2m/58.5)+(10.0\mathrm{g}-m)/180\}/\mathrm{g}{\mathrm{mol}}^{-1}}{0.250\mathrm{L}}$

Now $\mathrm{\Pi }=cRT$ we have

$\frac{15atm}{{\displaystyle \left(0.082\mathrm{LatmK}={\mathrm{mol}}^{-1}\right)\left(300\mathrm{K}\right)}}$$=\frac{\left\{\right(2m/58.5)+(10.0\mathrm{g}-m)/180\}/\mathrm{g}{\mathrm{mol}}^{-1}}{0.250\mathrm{L}}$

or $\left(\frac{2m}{58.5}+\frac{10.0g-m}{180}\right)/\mathrm{g}=\frac{15\times 0.250}{\left(0.082\right)\left(300\right)}=0.1524$

or $m=\frac{(180\times 0.1524-10.0)\left(58.5\right)}{(360-58.5)}\mathrm{g}=3.38\mathrm{g}$

Mass percentage of $\mathrm{NaCl}=\frac{3.38}{10}\times 100=33.8\mathrm{\%}$