Q. An 1% solution of KCl (I), NaCl (II), $BaCl_{2}\$(III) and urea (IV) have their osmotic pressure at the same temperature in the ascending order (molar masses of NaCl, KCl, $BaCl_{2}$ and urea are respectively 58.5, 74.5, 208.4 and 60 g $mol^{- 1}$). Assume 100% ionisation of the electrolytes at this temperature
Explanation
(d)
1% solution means 1 g solute is present in 100 mL of water.
Osmotic pressure, $\pi = \frac{iw \times RT}{M \times V}\$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \pi_{KCl} = \frac{2 \times 1 \times 1000 \times RT}{74.5 \times 100}$
=2$\times$0.134 RT
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \pi}_{NaCl} = \frac{2 \times 1 \times 1000 \times RT}{58.5 \times 100}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2 \times 0.171\ RT$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \pi_{BaCl} = \frac{3 \times 1 \times 1000 \times RT}{208.4 \times 100}\$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3 \times 0.048\ RT$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \pi_{urea} = \frac{1 \times 1 \times 1000 \times RT}{60 \times 100}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 \times 0.167\ RT\ \ \$
Since, temperature is same in all cases, the ascending order of osmotic pressure is
$III < \ IV < \ I\ < II$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.