Q.

Determine the osmotic pressure of the solution prepared by dissolving 25 mg of K2SO4 in 2L of water at 25oC. (Assuming it is to be completely dissociated)

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

5.27 x 10-4 atm

5.27 x 10-3 atm

5.27 x 10-2 atm

5.27 x 10-1 atm

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

K2SO4 dissolved = 25 mg = 0.025 gVolume = 2 L; Temperature, T = 298 KMolar mass of K2SO4 = 174 g mol-1K2SO4 dissociates into 2 K+ and SO42-ions, so i = 3π=iCRTπ=i×wM×RTV =3×0.025174×12×0.0821×298 =5.27×10−3atm

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET