Solutions and Colligative Properties - Result Question 1-1
1. A solution is prepared by dissolving $0.6$ g of urea (molar mass $=60 $ $\mathrm{g} $ $\mathrm{mol}^{-1}$ ) and $1.8$ g of glucose (molar mass $=180$ $ \mathrm{g}$ $\mathrm{mol}^{-1}$ ) in $100$ mL of water at $27^{\circ} \mathrm{C}$. The osmotic pressure of the solution is $( R = 0 .08206$ $ L $ $atm$ $ K^{-1} $ $mol^{-1})$
(2019 Main, 12 April II)
(a) $ 8.2$ atm
(b) $2.46$ atm
(c) $4.92$ atm
(d) $1.64$ atm
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Answer:
Correct Answer: 1. (c)
Solution:
Key Idea: Osmotic pressure is proportional to the molarity $(C)$ of the solution at a given temperature ( $T$ ).
Thus, $\pi \propto C, \pi=C R T$ (for dilute solution)
$ \pi=\frac{n}{V} R T $
For the relation, $\pi=C R T=\frac{n}{V} R T$
Given, mass of urea $=0.6 \mathrm{~g}$
Molar mass of urea $=60 \mathrm{~g} \mathrm{~mol}^{-1}$
Mass of glucose $=1.8 \mathrm{~g}$
Molar mass of glucose $=180 \mathrm{~g} \mathrm{~mol}^{-1}$
$ \begin{aligned} \pi & =\frac{\left.\left.\left[n_2 \text { (urea }\right)+n_2 \text { (glucose }\right)\right]}{V} R T \\ & =\frac{\left(\frac{0.6}{60}+\frac{1.8}{180}\right)}{100} \times 1000 \times 0.0821 \times 300 \\ & =(0.01+0.01) \times 10 \times 0.0821 \times 300 \\ \pi & =4.92 \mathrm{~atm} \end{aligned} $
BETA
