CameraIcon

SearchIcon

MyQuestionIcon

4

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Raoult's Law

The osmotic p...

Question

The osmotic pressure of $\frac{M}{20}$ solution of urea at $27^{o} C$ is :

A

12.315

atm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

1.2315

atm

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

0.12315

atm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

0.0123

atm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $1.2315$ atm
The temperature is $27^{o} C$ which is equal to $300 K$ .
Molarity, $M = \frac{1}{20} M$
Now,

Osmotic pressure, $Π = M R T$ $= \frac{1}{20} \times 0.0821 \times 300$ $= 1.2315 a t m$

Suggest Corrections

thumbs-up

1

Similar questions

Q. Find out the osmotic pressure of 0.25 M aqueous solution of urea at

27^{o} C

(R = 0.082 litre atm

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

Q. 100 mL aqueous solution of glucose with osmotic pressure 1.2 atm at

25^{o} C

is mixed with 300 mL aqueous solution of urea at 2.4 atm at

25^{o} C

. Calculate the osmotic pressure of mixture.

Q. If the osmotic pressure of a 0.010 M aqueous solution of sucrose at

27^{o} C

is 0.26 atm, then the osmotic pressure of a 0.010 M aqueous solution of NaCl at

27^{o} C

Q. Osmotic pressure of

0.01 M

aqueous urea is

0.24 a t m .

0.02 M

C a C l_{2} (a q)

is taken at same temperature, then

[Take : R = 0.08 a t m L m o l^{- 1} K^{- 1}]

What is the osmotic pressure of the solution obtained by mixing 300cc of 2% (m/v) solution of urea with 300cc of 3.42% solution of sucrose at 20^°C? (R = 0.0821 L atm K^-1 mol^-1)

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Liquids in Liquids and Raoult's Law

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon