Q. In an osmotic pressure measurement experiment, a 5% solution of compound ‘X’ is found to be isotonic with a 2 % acetic acid solution . The gram molecular mass of ‘X’ is
A) 24
B) 60
C) 150
D) 300
Explanation
(c)
2 % acetic acid solution
= $\frac{2 \times 1000}{60 \times 100}$M acetic acid
=0.33 M acetic acid
As the solution of compound “X” is isotonic to acetic acid solution, the molarity of solution of “X” will also be equal to 0.33 M. This is 5% solution. Hence
Mol.wt. of “X”=$\frac{5 \times 1000}{0.33 \times 100} = 150$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.