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Question: Answered & Verified by Expert
Which one of the following is the ratio of the lowering of vapour pressure of $0.1 \mathrm{M}$ aqueous solutions of $\mathrm{BaCl}_2, \mathrm{NaCl}$ and $\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$ respectively?
ChemistrySolutionsAP EAMCETAP EAMCET 2011
Options:
  • A 3 : 2 : 5
  • B 5 : 2 : 3
  • C 5 : 3 : 2
  • D 2 : 3 : 5
Solution:
2831 Upvotes Verified Answer
The correct answer is: 3 : 2 : 5
Lowering of vapour pressure is a colligative property. It depends on the number of particles of the solute.
$\begin{aligned}
\mathrm{BaCl}_2 & \rightleftharpoons \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-} ; 3 \text { ions } \\
\mathrm{NaCl} & \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-} ; 2 \text { ions } \\
\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 & \rightleftharpoons 2 \mathrm{Al}^{3+}+3 \mathrm{SO}_4^{3-} ; 5 \text { ions }
\end{aligned}$
Hence the ratio of lowering of vapour pressure is $3: 2: 5$.

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