A closed container contains mixture of non-reacting gases A and B. Partial pressure of $A$ and $B$ are 4.5…

A closed container contains mixture of non-reacting gases A and B. Partial pressure of $A$ and $B$ are 4.5 bar and 5.5 . bar respectively. Find mole fraction of $\mathrm{A}$ and $\mathrm{B}$ respectively.
  1. 0.035 and 0.065
  2. 0.055 and 0.045
  3. 0.45 and 0.55
  4. 0.55 and 0.45

Solution

$\begin{aligned} \mathrm{P}_{\text {Total }}=\mathrm{P}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}} & =4.5+5.5 \\ & =10.0 \mathrm{bar}\end{aligned}$ $\mathrm{P}_{\mathrm{A}}=x_{\mathrm{A}} \times \mathrm{P}_{\text {Total }}$ $x_{\mathrm{A}}=\frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\text {Total }}}=\frac{4.5}{10}=0.45$ $x_{\mathrm{B}}=\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\text {Total }}}=\frac{5.5}{10}=0.55$

Asked in: MHT CET 2023 (13 May Shift 2)

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