If 20 g of a solute was dissolved in 500 mL of water and osmotic pressure of the solution was found to be 600 mm of Hg at 15° C, then molecular weight of the solute is:

Question

If 20 g of a solute was dissolved in 500 mL of water and osmotic pressure of the solution was found to be 600 mm of Hg at 15° C, then molecular weight of the solute is: (A) 1000 (B) 1200 (C) 1400 (D) 1800

Tardigrade Admin · Accepted Answer

Osmotic pressure is colligative property. Use the formula

πV = n RT

to solve problem.

πV = n RT

or

π = \frac{n}{V} RT = \frac{m RT}{M V}

where

π =

osmotic presure

= 600 mm

of

H g

= 600/760 a t m

m =

mass of solute

= 20 g

M =

molecular mass of solute

R = 0.0821 L a t m m o l^{- 1} K^{- 1}

T = 1 5^{\circ} C = 15 + 273 = 288 K

V = 500 m L = 0.5 L

\frac{600}{760} = \frac{20 \times 0.0821 \times 288}{M \times 0.5}

or

M = 1200

∴

molecular mass of solute is

1200

.

Q. If 20g of a solute was dissolved in 500mL of water and osmotic pressure of the solution was found to be 600mm of Hg at 15∘C, then molecular weight of the solute is:

1000

1200

1400

1800

Q. If $20 g$ of a solute was dissolved in $500 m L$ of water and osmotic pressure of the solution was found to be $600 mm$ of $H g$ at $1 5^{\circ} C$ , then molecular weight of the solute is: