The osmotic pressure of a solution (in bar) obtained by dissolving 0.35 g of CaCl2 (van’t Hoff factor, i = 2.47) in 3.5 L of water at 345 K will be approximately
[Given : Molar mass of CaCl2 = 110.9 g mol−1 and R = 0.083 L bar mol−1 K−1]
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.064
The osmotic pressure of a dilute solution is given by the van’t Hoff equation. For an electrolyte, the van’t Hoff factor i corrects for the extra particles formed on dissociation:
\( \displaystyle \pi=iCRT=i\,\frac{n}{V}\,RT \)
Moles of \( \ce{CaCl2} \):
\( \displaystyle n=\frac{0.35\ \mathrm{g}}{110.9\ \mathrm{g\,mol^{-1}}}=3.16\times10^{-3}\ \mathrm{mol} \)
Osmotic pressure (V = 3.5 L, T = 345 K):
\( \displaystyle \pi=2.47\times\frac{3.16\times10^{-3}}{3.5}\times0.083\times345 \)
\( \displaystyle \pi=2.47\times(9.02\times10^{-4})\times28.64 \)
\( \displaystyle \pi=0.0638\ \mathrm{bar}\approx0.064\ \mathrm{bar} \)
So the osmotic pressure is about 0.064 bar.