A solution is prepared by dissolving 0.3 g of a nonvolatile non-electrolyte solute ' A ' of molar mass $60 \mathrm{~g} \mathrm{~mol}^{-1}$ and 0.9 g of a non-volatile nonelectrolyte solute ' B ' of molar mass $180 \mathrm{~g} \mathrm{~mol}^{-1}$ in $100 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}$ at $27^{\circ} \mathrm{C}$. Osmotic pressure of the solution will be [Given : $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Solutions

A solution is prepared by dissolving 0.3 g of a nonvolatile non-electrolyte solute ' A ' of molar mass

60 g mol^{- 1}

and 0.9 g of a non-volatile nonelectrolyte solute ' B ' of molar mass

180 g mol^{- 1}

100 mL H_{2} O

2 7^{\circ} C

. Osmotic pressure of the solution will be
[Given :

R = 0.082 L atm K^{- 1} mol^{- 1}

]

(A)1.23 atm
(B)2.46 atm✓ Correct
(C)0.82 atm
(D)1.47 atm

Answer: (B)

Solution

The two non-electrolyte solutes give

n_{A} = \frac{0.3}{60} = 0.005 mol

and

n_{B} = \frac{0.9}{180} = 0.005 mol

, a total of

0.01 mol

. In

0.1 L

of solution the concentration is

0.1 M

. The osmotic pressure is

π = C R T = 0.1 \times 0.082 \times 300 = 2.46 atm

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option

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