Question
A solution is prepared by dissolving 0.3 g of a nonvolatile non-electrolyte solute ' A ' of molar mass $60\text{ }g{\text{ }mol}^{- 1}$ and 0.9 g of a non-volatile nonelectrolyte solute ' B ' of molar mass $180\text{ }g{\text{ }mol}^{- 1}$ in $100\text{ }mLH_{2}O$ at $27^{\circ}C$. Osmotic pressure of the solution will be
[Given: $R = 0.082\text{ }L\text{ }atm{\text{ }K}^{- 1}{\text{ }mol}^{- 1}$ ]
Options
Solution
Mass of solute ' A ' $= 0.3\text{ }g$
Moles of solute ' A ' $= \frac{0.3\text{ }g}{60\text{ }g/mol} = \frac{1}{200}\text{ }mol$
Mass of solute $\ 'B' = 0.9\text{ }g$
Moles of solute $\ 'B' = \frac{0.9gm}{180\text{ }g/mol} = \frac{1}{200}\text{ }mol$
Total molarity of all solutes
$$\begin{matrix} & \ = \frac{2/200}{100} \times 1000 = \frac{1}{10}M \\ \therefore\pi & \ = \frac{1}{10} \times 0.082 \times 300 \\ \pi & \ = 2.46\text{ }atm. \end{matrix}$$
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