18 g of glucose (molar mass = 180 g/mol) is dissolved in water to prepare 500 ml solution at 1 5 ∘ C . Calculate the osmotic pressure of the solution. [ R = 0.0821 L atm K − 1 mol − 1 ]

Number of moles of glucose, n = 180 18 = 0.1 mol Volume of solution, V = 500 ml = 0.5 L Temperature, T = 15 + 273 = 288 K Osmotic pressure is given by π = CRT = V n RT Substituting the values: π = 0.5 0.1 × 0.0821 × 288 π = 0.2 × 0.0821 × 288 π = 4.72896 atm ≈ 4.73 atm Answer: 4.73 atm

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ChemistryLiquid SolutionAbnormal Colligatiove PropertiesMedium2 minPYQ_2026

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18 g of glucose (molar mass = 180 g/mol) is dissolved in water to prepare 500 ml solution at $1 5^{\circ} C$ . Calculate the osmotic pressure of the solution. [ $R = 0.0821 L atm K^{- 1} mol^{- 1}$ ]

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Answer:

Solution:

Number of moles of glucose, $n = \frac{18}{180} = 0.1 mol$

Volume of solution, $V = 500 ml = 0.5 L$

Temperature, $T = 15 + 273 = 288 K$

Osmotic pressure is given by $π = CRT = \frac{n}{V} RT$

Substituting the values:

$π = \frac{0.1}{0.5} \times 0.0821 \times 288$

$π = 0.2 \times 0.0821 \times 288$

$π = 4.72896 atm \approx 4.73 atm$

Answer: $4.73 atm$

Stream:MHT_CETSubject:ChemistryTopic:Liquid SolutionSubtopic:Abnormal Colligatiove Properties

⏱ 2mℹ️ Source: PYQ_2026

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