Question

An 1% solution of $ KCl $ (I), $ NaCl $ (II), $ BaC{{l}_{2}} $ (III) and urea (IV) have their osmotic pressure at the same temperature in the ascending order (molar masses of $ NaCl,KCl,BaC{{l}_{2}} $ and urea are respectively $58.5, 74.5, 208.4$ and $60\, g$. $ mo{{l}^{-1}}) $ .Assume $100\%$ ionization of the electrolytes at this temperature

• A. $ I < III < II < IV $
• B. $ III < I < II < IV $
• C. $ I < II < III < IV $
• D. $ I < III < IV < II $
• E. $ III < IV < I < II $

Answer 1

Answer: (E)

1% solution means 1 g solute is present in 100 mL of water. Osmotic pressure,
$ \pi =\frac{iw\times RT}{M\times V} $
$ \therefore $ $ {{\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 }_{BaC{{l}_{2}}}}=\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 $