Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

50 mL of 0.2 molal urea solution (density = 1.012 g mL–1 at 300 K) is mixed with 250 mL of a solution containing 0.06 g of urea. Both the solutions were prepared in the same solvent. The osmotic pressure (in torr) of the resulting at 300 K is ____.
[Use : Molar mass of urea = 60 g mol–1; gas constant, R = 62 L torr K–1 mol–1;
Assume, mix H = 0, mix V = 0]

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

answer is 682.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

detailed_solution_thumbnail

solution-I 50ml, 0.2 molal urea solution with density 1.012 g/L of solution wt of solution =50.6 gram Let urea = x gram 
Solvent = (50.6-x) gram  
Molality = x×100060×(50.6x)
0.2=1000x60×50.6x
X=0.6gram
Solution II-  0.06 gram urea in 250ml. solution
Total wt of urea =0.6+0.06
 = 0.66gram
Total volume =50+250  =300ml
π=C.RT=0.66×100060×300×62×300=682 torr 

Watch 3-min video & get full concept clarity

courses

No courses found

Ready to Test Your Skills?

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

Take Free Test

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon