Osmotic pressure of a solution is 0.0821 atm at a temperature of 300 K. The concentration in .mole per litre will be:

Question

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

Tardigrade Admin · Accepted Answer

V=n.RT or (n/V)=(π /RT) =(0.0821/300× 0.0821) =0.3× 10-2 mole/litre

Q. Osmotic pressure of a solution is $0.0821$ atm at a temperature of $300 K$ . The concentration in .mole per litre will be:

$0.33$

$0.066$

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

$3$

$V = n . RT$ or $\frac{n}{V} = \frac{π}{RT}$
$= \frac{0.0821}{300 \times 0.0821}$
$= 0.3 \times 1 0^{- 2}$ mole/litre

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

0.33

0.066

0.3×10−2

3

V=n.RT or Vn​=RTπ​ =300×0.08210.0821​ =0.3×10−2 mole/litre

Q. Osmotic pressure of a solution is $0.0821$ atm at a temperature of $300 K$ . The concentration in .mole per litre will be:

$0.33$

$0.066$

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

$3$

$V = n . RT$ or $\frac{n}{V} = \frac{π}{RT}$
$= \frac{0.0821}{300 \times 0.0821}$
$= 0.3 \times 1 0^{- 2}$ mole/litre