Solutions Question 13
Question 13 - 31 January - Shift 1
At $27^{\circ} C$, a solution containing $2.5 g$ of solute in $250.0 mL$ of solution exerts an osmotic pressure of $400 Pa$. The molar mass of the solute is g $mol^{-1}$ (Nearest integer)
(Given : $R=0.083 L bar K^{-1} mol^{-1}$ )
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Answer: (62250)
Solution:
: $\pi=CRT$
$ \frac{400 Pa}{10^{5}}=\frac{\frac{2.5 g}{M_o}}{250 / 1000 L} \times 0.83 \frac{L-bar}{K \cdot mol} \times 300 K $