The Question is from neet-chemistry , and it was created by the author THIRUPATHI REDDY

Question

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 $({P_2})$ and 10 g of sucrose $\left( {{C_{12}}{H_{12}}{O_{11}}} \right)$ in 250 ml of water $({P_3}).$ The right option for the decreasing order of osmotic pressure of these solutions is:

THIRUPATHI REDDY · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Osmotic Pressure
	**Related Topics**: Colligative Properties, Molarity, Solution Chemistry
	**Prerequisites**: Understanding of osmotic pressure, molar mass calculations, and the relationship between concentration and osmotic pressure.
## SOLUTION PATHWAY
**Given Information**:
	Mass of each solute = 10 g
	Volume of solvent (water) = 250 ml = 0.25 L
**Method & Steps**:
1. Calculate the number of moles of each solute:
	Reasoning: Osmotic pressure is directly proportional to the molar concentration of the solute.
	Formula/Rule: $n = \frac{mass}{Molar\ mass}$
	Glucose ($C_6H_{12}O_6$): $n_1 = \frac{10\ g}{180\ g/mol} \approx 0.0556\ mol$
	Urea ($CH_4N_2O$): $n_2 = \frac{10\ g}{60\ g/mol} \approx 0.1667\ mol$
	Sucrose ($C_{12}H_{22}O_{11}$): $n_3 = \frac{10\ g}{342\ g/mol} \approx 0.0292\ mol$
2. Calculate the molarity of each solution:
	Working: $Molarity (M) = \frac{moles\ of\ solute}{volume\ of\ solution\ in\ liters}$
	$M_1 = \frac{0.0556\ mol}{0.25\ L} \approx 0.222\ M$
	$M_2 = \frac{0.1667\ mol}{0.25\ L} \approx 0.667\ M$
	$M_3 = \frac{0.0292\ mol}{0.25\ L} \approx 0.117\ M$
3. Compare the molarities to determine the order of osmotic pressure:
	Result: $M_2 > M_1 > M_3$, therefore $P_2 > P_1 > P_3$
	Verification: Since osmotic pressure is directly proportional to molarity, the solution with the highest molarity will have the highest osmotic pressure.
## OPTION ANALYSIS
[A] ✓: $P_2 > P_1 > P_3$. This is correct as Urea has the highest molarity, followed by glucose and then sucrose.
[B] :x: $P_1 > P_2 > P_3$. Incorrect order of molarity.
[C] :x: $P_3 > P_1 > P_2$. Incorrect order of molarity.
[D] :x: $P_1 > P_3 > P_2$. Incorrect order of molarity.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrect calculation of molar mass: Double-check the molar mass calculations for each solute.
		2. Confusing mass and moles: Remember to convert mass to moles before calculating molarity.
	**Quick Tips**:
		1. For the same mass of solute, the compound with the lowest molar mass will have the highest molarity and thus the highest osmotic pressure.
		2. Osmotic pressure is a colligative property, meaning it depends on the number of solute particles, not their identity.

THIRUPATHI REDDY · Answer

The correct Option is "P₂ > P₁ > P₃"

THIRUPATHI REDDY · Answer

P₁ > P₂ > P₃

THIRUPATHI REDDY · Answer

P₃ > P₁ > P₂

THIRUPATHI REDDY · Answer

P₁ > P₃ > P₂

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