Find the osmotic pressure of 10\% (w/v) solution of sucrose at 27^ ^ C . (Molar mass of sucrose = 342 g/mol)

Correct answer: C. 7.20 atm Given concentration of sucrose = 10\% (w/v) This means 10 g of sucrose is present in 100 mL of solution. Mass of sucrose, W = 10 g Volume of solution, V = 100 mL = 0.1 L Molar mass of sucrose, M = 342 g/mol Temperature, T = 27^ ^ C = 300 K Gas constant, R = 0.0821 L atm K^ -1 mol^ -1 Osmotic pressure, π = CRT = (n)/(V)RT π = (W)/(M × V)RT π = (10)/(342 × 0.1) × 0.0821 × 300 π = (100)/(342) × 24.63 π = 7.20 atm Answer: 7.20 atm

MHT CET 202626 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 26 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Solutions

MHT CET 2026 · 26 April · Morning Shift · Chemistry · Solutions

Find the osmotic pressure of 10\% (w/v) solution of sucrose at 27^...

Find the osmotic pressure of

10\%

(w/v) solution of sucrose at

27^{}^{\circ}\text{C}

. (Molar mass of sucrose

= 342

g/mol)

Options

A $3.46$ atm
B $5.87$ atm
C $7.20$ atm
D $8.79$ atm

Correct answer

C. $7.20$ atm

Step-by-step solution

Given concentration of sucrose

= 10\%

(w/v)

This means

10\text{ g}

of sucrose is present in

100\text{ mL}

of solution.

Mass of sucrose,

W = 10\text{ g}

Volume of solution,

V = 100\text{ mL} = 0.1\text{ L}

Molar mass of sucrose,

M = 342\text{ g/mol}

Temperature,

T = 27^{}^{\circ}\text{C} = 300\text{ K}

Gas constant,

R = 0.0821\text{ L atm K}^{-1}\text{ mol}^{-1}

Osmotic pressure,

\pi = CRT = \dfrac{n}{V}RT

\pi = \dfrac{W}{M \times V}RT

\pi = \dfrac{10}{342 \times 0.1} \times 0.0821 \times 300

\pi = \dfrac{100}{342} \times 24.63

\pi = 7.20\text{ atm}

Answer:

7.20

atm

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