The following solutions were prepared by dissolving 10 g of glucose ($C_6H_{12}O_6$) in 250 ml of water ($P_1$

Question

The following solutions were prepared by dissolving 10 g of glucose ($C_6H_{12}O_6$) in 250 ml of water ($P_1$), 10 g of urea ($CH_4N_2O$) 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 decreasing order of osmotic pressures of these solutions is:

Accepted Answer

Osmotic pressure is a colligative property and for dilute solutions, it is given by the formula $\Pi = \frac{w_2 RT}{M_2 V}$ . Here, the mass of solute ($w_2 = 10 	ext{ g}$) and the volume of the solution ($V = 250 	ext{ mL}$) are constant for all three cases. Therefore, the osmotic pressure is inversely proportional to the molar mass of the solute ($\Pi \propto \frac{1}{M_2}$).

Let's calculate the molar mass for each solute:
1. Glucose ($C_6H_{12}O_6$): $M_1 = 180 	ext{ g mol}^{-1}$
2. Urea ($CH_4N_2O$): $M_2 = 60 	ext{ g mol}^{-1}$
3. Sucrose ($C_{12}H_{22}O_{11}$): $M_3 = 342 	ext{ g mol}^{-1}$

The order of molar masses is: Sucrose ($M_3$) > Glucose ($M_1$) > Urea ($M_2$).
Thus, the decreasing order of their osmotic pressures will be the reverse of the order of their molar masses:
$\Pi(	ext{Urea}) > \Pi(	ext{Glucose}) > \Pi(	ext{Sucrose})$, which is $P_2 > P_1 > P_3$.

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