An aqueous solution of a substance with molar mass 240 has osmotic pressure 0.2 atm at 300 K. The density of solution in g/dm3 is: (R = 0.08 litre atm K-1 mol-1)

Question

An aqueous solution of a substance with molar mass 240 has osmotic pressure 0.2 atm at 300 K. The density of solution in g/dm3 is: (R = 0.08 litre atm K-1 mol-1) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

π=(n/V)RT π=(w/MV)RT π=(dRT/M) ⇒ d=(π M/RT) =(0.2 × 240/0.08 × 300)=2

Q. An aqueous solution of a substance with molar mass $240$ has osmotic pressure $0.2$ atm at $300 K$ . The density of solution in $g / d m^{3}$ is: $(R = 0.08$ litre atm $K^{- 1} m o l^{- 1})$

$π = \frac{n}{V} RT$
$π = \frac{w}{M V} RT$
$π = \frac{d RT}{M}$
$\Rightarrow d = \frac{π M}{RT}$
$= \frac{0.2 \times 240}{0.08 \times 300} = 2$

Q. An aqueous solution of a substance with molar mass 240 has osmotic pressure 0.2 atm at 300K. The density of solution in g/dm3 is: (R=0.08 litre atm K−1mol−1)

π=Vn​RT π=MVw​RT π=MdRT​ ⇒d=RTπM​ =0.08×3000.2×240​=2

Q. An aqueous solution of a substance with molar mass $240$ has osmotic pressure $0.2$ atm at $300 K$ . The density of solution in $g / d m^{3}$ is: $(R = 0.08$ litre atm $K^{- 1} m o l^{- 1})$

$π = \frac{n}{V} RT$
$π = \frac{w}{M V} RT$
$π = \frac{d RT}{M}$
$\Rightarrow d = \frac{π M}{RT}$
$= \frac{0.2 \times 240}{0.08 \times 300} = 2$