The following solutions were prepared by dissolving 10 g of glucose in 250 mL of water (), 10 g of urea in 250 mL of water () and 10 g of sucrose in 250 mL of water (). The right option for the decreasing order of osmotic pressure of these solutions is

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Solution

Q.
The following solutions were prepared by dissolving 10 g of glucose $(C_{6} H_{12} O_{6})$ in 250 mL of water ( $P_{1}$ ), 10 g of urea $(C H_{4} N_{2} O)$ in 250 mL of water ( $P_{2}$ ) and 10 g of sucrose $(C_{12} H_{22} O_{11})$ in 250 mL of water ( $P_{3}$ ). The right option for the decreasing order of osmotic pressure of these solutions is [2021]

1 $P_{3} > P_{1} > P_{2}$

2 $P_{2} > P_{1} > P_{3}$

3 $P_{1} > P_{2} > P_{3}$

4 $P_{2} > P_{3} > P_{1}$

Ans.
(2)

Osmotic pressure $(π) = C R T$

$∴$ $π \propto C$

For glucose solution, $C_{1} = \frac{10}{180} \times \frac{1000}{250} = 0.22$ $M$

For urea solution, $C_{2} = \frac{10}{60} \times \frac{1000}{250} = 0.66$ $M$

For sucrose solution, $C_{3} = \frac{10}{342} \times \frac{1000}{250} = 0.117$ $M$

Hence, order of osmotic pressure is $P_{2} > P_{1} > P_{3}$

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