Question

The 'X' g nonvolatile solute having molar mass 196 g mol^-1 is dissolved in 3 dm³ water. Calculate the value of 'X' if resulting solution has osmotic pressure 0.8 atm at 300 К.
(R = 0.0821dm³ atm K^-1 mol^-1)

Options

• 19.1g

• 9.8 g

• 14.7 g

• 14.7 g

Answer 1

19.1g

Explanation:

\[\pi=\frac{\mathrm{W}_2\mathrm{RT}}{\mathrm{M}_2\mathrm{V}}\]
\[\therefore\quad\mathrm{W}_2=\frac{\pi\mathrm{~M}_2\mathrm{V}}{\mathrm{RT}}\]
\[\begin{aligned} & =\frac{0.8\mathrm{atm}\times196\mathrm{gmol}^{-1}\times3\mathrm{dm}^3}{0.0821\mathrm{atm}\mathrm{dm}^3\mathrm{K}^{-1}\mathrm{mol}^{-1}\times300\mathrm{K}} \\ & =19.098\mathrm{g}\approx19.1\mathrm{g} \end{aligned}\]