JEE MAIN - Chemistry (2007 - No. 30)
A 5.25 % solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol−1) in
the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm−3, molar mass of the substance will be
90.0 g mol−1
115.0 g mol−1
105.0 g mol−1
210.0 g mol−1
Explanation
TIPS/FORMULAE:
Osmotic pressure $$\left( \pi \right)$$ of isotonic solutions are equal.
For solution of unknown substance $$\left( {\pi = CRT} \right)$$
$${C_1} = {{5.25/M} \over V}$$
For solution of urea,
$$C{}_2\,\,$$ (concentration) $$ = {{1.5/60} \over V}$$
Given, $${\pi _1} = {\pi _2}$$
as $$\,\,\,\,\pi = CRT$$
$$\therefore$$ $$\,\,\,\,\,{C_1}RT = {C_2}RT\,\,$$ or $$\,\,{C_1} = {C_2}$$
or $$\,\,\,\,{{5.25/M} \over V} = {{1.8/60} \over V}$$
$$\therefore$$ $$\,\,\,\,\,M = 210g/mol$$
Osmotic pressure $$\left( \pi \right)$$ of isotonic solutions are equal.
For solution of unknown substance $$\left( {\pi = CRT} \right)$$
$${C_1} = {{5.25/M} \over V}$$
For solution of urea,
$$C{}_2\,\,$$ (concentration) $$ = {{1.5/60} \over V}$$
Given, $${\pi _1} = {\pi _2}$$
as $$\,\,\,\,\pi = CRT$$
$$\therefore$$ $$\,\,\,\,\,{C_1}RT = {C_2}RT\,\,$$ or $$\,\,{C_1} = {C_2}$$
or $$\,\,\,\,{{5.25/M} \over V} = {{1.8/60} \over V}$$
$$\therefore$$ $$\,\,\,\,\,M = 210g/mol$$
Comments (0)
