A $5.13\mathrm{\%}$ solution of sucrose $\left({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}\right)$ is isotonic with a $0.9\mathrm{\%}$ solution of an unknown solute. Calculate the molar mass of the solute. 

For sucrose, ${\pi }_1=?;$$\mathrm{Mol}.$ w.t. of ${\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}$, $M_2=(12\times 12)+(22\times 1)+(11\times 16)=342\mathrm{g}{\mathrm{mol}}^{-}$, wt. of solute $W_2=5.13\mathrm{g};$ concentration ${\mathrm{C}}_1=\frac{{\mathrm{W}}_1}{{\mathrm{M}}_1}$

volume of solution, $V=100\mathrm{mL}\times \frac{1\mathrm{L}}{1000\mathrm{mL}}=0.1\mathrm{L}$

$\mathbf{\therefore }{\mathrm{\pi }}_1=\frac{{\mathrm{W}}_1}{{\mathrm{M}}_1\mathrm{V}}\mathrm{RT}=\frac{5.13\mathrm{gRT}}{342{\mathrm{gmol}}^{-}\times 0.1\mathrm{L}}$       ….(1)

(ii) For unknown solute ${\pi }_2=?;\text{ mol }$wt. of solute,$M_2=?;$ wt. of solute $W_2=0.9\mathrm{g}$;

Volume $=100\mathrm{mL}\times \frac{1\mathrm{L}}{1000\mathrm{mL}}=0.1\mathrm{L}$ we know that,

${\pi }_2=\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2\mathrm{V}}\times RT=\frac{0.9\mathrm{g}}{{\mathrm{M}}_2\times 0.1\mathrm{L}}\times RT$     ….. (2)

Since the solutions are isotonic, their osmotic pressures are equal. So. ${\pi }_1={\pi }_2$ Hence from equations (1) and (2), we have :

$\frac{5.13\mathrm{g}\times RT}{342\mathrm{g}{\mathrm{mol}}^{-}\times 0.1\mathrm{L}}=\frac{0.9\mathrm{g}\times RT}{M_2\times 0.1\mathrm{L}}$

$\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }{M}_2=\frac{0.9\mathrm{g}\times RT\times 342\mathrm{g}{\mathrm{mol}}^{-}\times 0.1\mathrm{L}}{0.1\mathrm{L}\times 5.13\mathrm{g}\times RT}$

$\mathbf{\therefore }{\mathit{M}}_{\mathbf{2}}\mathbf{=}\mathbf{60}\mathbf{g}{\mathbf{mol}}^{\mathbf{-}}$