JEE MAIN - Chemistry (2023 - 31st January Morning Shift - No. 20)
At $$27^{\circ} \mathrm{C}$$, a solution containing $$2.5 \mathrm{~g}$$ of solute in $$250.0 \mathrm{~mL}$$ of solution exerts an osmotic pressure of $$400 \mathrm{~Pa}$$. The molar mass of the solute is ___________ $$\mathrm{g} \mathrm{~mol}^{-1}$$ (Nearest integer)
(Given : $$\mathrm{R}=0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$$)
Answer
62250
Explanation
$$\pi = CRT$$
$$ \begin{aligned} 400 & =\frac{2.5}{\mathrm{m_w}} \times 4 \times\left(.083 \times 10^5\right) \times 300 \\\\ \mathrm{m_w} & =\frac{10 \times 0.083 \times 3}{4} \times 10^5 \\\\ & =62250 \end{aligned} $$
$$ \begin{aligned} 400 & =\frac{2.5}{\mathrm{m_w}} \times 4 \times\left(.083 \times 10^5\right) \times 300 \\\\ \mathrm{m_w} & =\frac{10 \times 0.083 \times 3}{4} \times 10^5 \\\\ & =62250 \end{aligned} $$
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