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The partial pressure of ethane over a solution containing \(6.56 \times 10^{-3} \mathrm{~g}\) of ethane is 1 bar. If the solution contains \(\mathbf{5 \cdot 0 0} \times 10^{-2} \mathrm{~g}\) of ethane, then what shall be the partial pressure of the gas?
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We know that, \(m=K_H \times P\)
\(\begin{aligned} \therefore & 6 \cdot 56 \times 10^{-3} \mathrm{~g}=K_H \times 1 \text { bar } ...(i) \\ \therefore & 5 \cdot 00 \times 10^{-2} \mathrm{~g}=K_H \times P ...(ii)\\ & K_H=6.56 \times 10^{-3 / 1} \text { bar (from i) } \\ & K_H=5.00 \times 10^{-2} / p \text { bar (from ii) } \\ \therefore & \frac{6.56 \times 10^{-3}}{1}=\frac{5.00 \times 10^{-2}}{p} \\ \therefore & p=\frac{5.00 \times 10^{-2}}{6 \cdot 56 \times 10^{-3}}=0.762 \text { bar } \end{aligned}\)
\(\begin{aligned} \therefore & 6 \cdot 56 \times 10^{-3} \mathrm{~g}=K_H \times 1 \text { bar } ...(i) \\ \therefore & 5 \cdot 00 \times 10^{-2} \mathrm{~g}=K_H \times P ...(ii)\\ & K_H=6.56 \times 10^{-3 / 1} \text { bar (from i) } \\ & K_H=5.00 \times 10^{-2} / p \text { bar (from ii) } \\ \therefore & \frac{6.56 \times 10^{-3}}{1}=\frac{5.00 \times 10^{-2}}{p} \\ \therefore & p=\frac{5.00 \times 10^{-2}}{6 \cdot 56 \times 10^{-3}}=0.762 \text { bar } \end{aligned}\)
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