Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

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:

$P_2 > P_3 > P_1$

$P_3 > P_1 > P_2$

$P_2 > P_1 > P_3$

$P_1 > P_2 > P_3$

Step-by-Step Solution

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 ( $w_2 = 10 \text{ g}$ ) and the volume of the solution ( $V = 250 \text{ mL}$ ) 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:

Glucose ( $C_6H_{12}O_6$ ): $M_1 = 180 \text{ g mol}^{-1}$
Urea ( $CH_4N_2O$ ): $M_2 = 60 \text{ g mol}^{-1}$
Sucrose ( $C_{12}H_{22}O_{11}$ ): $M_3 = 342 \text{ g mol}^{-1}$

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: $\Pi(\text{Urea}) > \Pi(\text{Glucose}) > \Pi(\text{Sucrose})$ , which is $P_2 > P_1 > P_3$ .

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