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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.
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The correct answer is:
0.45 and 0.55
$\begin{aligned} \mathrm{P}_{\text {Total }}=\mathrm{P}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}} & =4.5+5.5 \\ & =10.0 \mathrm{bar}\end{aligned}$
$\mathrm{P}_{\mathrm{A}}=x_{\mathrm{A}} \times \mathrm{P}_{\text {Total }}$
$x_{\mathrm{A}}=\frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\text {Total }}}=\frac{4.5}{10}=0.45$
$x_{\mathrm{B}}=\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\text {Total }}}=\frac{5.5}{10}=0.55$
$\mathrm{P}_{\mathrm{A}}=x_{\mathrm{A}} \times \mathrm{P}_{\text {Total }}$
$x_{\mathrm{A}}=\frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\text {Total }}}=\frac{4.5}{10}=0.45$
$x_{\mathrm{B}}=\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\text {Total }}}=\frac{5.5}{10}=0.55$
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