MyQuestionIcon

11

You visited us 11 times! Enjoying our articles? Unlock Full Access!

Osmotic Pressure

Osmotic press...

Question

Osmotic pressure of a solution is 0.0821 atm at a temperature of 400 K. Calculate the concentration of solution in mol/litre.
$[R = 0.0821 L a t m K^{- 1} m o l^{- 1}]$

Open in App

Solution

The osmotic pressure
$π = C R T$
$0.0821 a t m = C \times 0.0821 L a t m / K m o l \times 400 K$
$C = 0.0025 m o l / L$
Hence, the concentration of solution is 0.0025 mol/L.

Suggest Corrections

thumbs-up

1

Similar questions

Q. A decimolar solution of

K_{4} [F e {(C N)}_{6}]

50 %

300

K. Calculate the osmotic pressure of solution.

(R= 0.0821 litre atm/K mol)

Q. Calculate the osmotic pressure of

0.5 %

solution of glucose (molecular mass 180) at

18^{o} C

. The value of solution constant is 0.0821

l i t r e - a t m K^{- 1} m o l^{- 1}

.

Q. (i) Prove that osmotic pressure is a colligative property.
(ii) Calculate the molar concentration of urea solution if it exerts an osmotic pressure of 2.45 atmosphere at 300K. [R= 0.0821 l. atm.

m o l^{1}

K^{- 1}

Q. Calculate the osmotic pressure of

5 %

solution of glucose at

25^{o} C

. (Molecular weight of Glucose

= 180

R = 0.0821

K^{- 1} m o l^{- 1}

Osmotic pressure is 0.0821 atm at a temperature of 300 K. Find concentration in mole/litre.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Osmotic Pressure

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Osmotic Pressure

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon