A solution is prepared by dissolving 0.6 g of urea (molar mass $=60 \text{ g mol}^{-1}$) and 1.8 g of glucose (molar mass $=180 \text{ g mol}^{-1}$) in 100 mL of water at $27^{\circ}\text{C}$. The osmotic pressure of the solution is : ($R=0.08206 \text{ L atm K}^{-1}\text{mol}^{-1}$)

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Solutions

A solution is prepared by dissolving 0.6 g of urea (molar mass

= 60 g mol^{- 1}

) and 1.8 g of glucose (molar mass

= 180 g mol^{- 1}

) in 100 mL of water at

2 7^{\circ} C

. The osmotic pressure of the solution is :
(

R = 0.08206 L atm K^{- 1} mol^{- 1}

)

(A)8.2 atm
(B)1.64 atm
(C)4.92 atm✓ Correct
(D)2.46 atm

Answer: (C)

Solution

Both solutes are non-electrolytes, so just add their moles: urea gives

\frac{0.6}{60} = 0.01

mol and glucose

\frac{1.8}{180} = 0.01

mol, a total of

0.02

mol in

0.1

L, so

C = 0.2

M. At

T = 300

π = C R T = 0.2 \times 0.08206 \times 300 \approx 4.92

atm.

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option

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