Osmotic pressure of a solution is 0.0821 atm at a temperature of 300 K. The concentration in moles/litre will be

Question

Osmotic pressure of a solution is 0.0821 atm at a temperature of 300 K. The concentration in moles/litre will be (A) 0.33 (B) 0.066 (C) 0.3 × 10-2 (D) 3

Tardigrade Admin · Accepted Answer

π = CRT C =(π/R T)=(0.0821/0.0821 × 300) =0.3 × 10-2 mol L -1

Q. Osmotic pressure of a solution is $0.0821 a t m$ at a temperature of $300 K$ . The concentration in moles/litre will be

0.33

0.066

$0.3 \times 1 0^{- 2}$

3

$π = CRT$
$C = \frac{π}{RT} = \frac{0.0821}{0.0821 \times 300}$
$= 0.3 \times 1 0^{- 2} m o l L^{- 1}$

Q. Osmotic pressure of a solution is 0.0821atm at a temperature of 300K. The concentration in moles/litre will be

0.33

0.066

0.3×10−2

3

π=CRT C=RTπ​=0.0821×3000.0821​ =0.3×10−2molL−1

Q. Osmotic pressure of a solution is $0.0821 a t m$ at a temperature of $300 K$ . The concentration in moles/litre will be

$0.3 \times 1 0^{- 2}$

$π = CRT$
$C = \frac{π}{RT} = \frac{0.0821}{0.0821 \times 300}$
$= 0.3 \times 1 0^{- 2} m o l L^{- 1}$