The partial pressure of ethane over a solutio | vedclass

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(A) According to Henry's law,the partial pressure of a gas is directly proportional to its mole fraction in the solution: $p = K_H \cdot x$.Since the

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$0.764$

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(A) According to Henry's law,the partial pressure of a gas is directly proportional to its mole fraction in the solution: $p = K_H \cdot x$.Since the amount of solute is very small compared to the solvent,the mole fraction $x$ is approximately equal to the ratio of the number of moles of solute to the number of moles of solvent $(n_{solute} / n_{solvent})$.Thus,$p \propto n_{solute}$.Given $p_1 = 1 \ bar$ for $n_1 = \frac{6.56 	imes 10^{-2} \ g}{30 \ g \ mol^{-1}}$.We need to find $p_2$ for $n_2 = \frac{5.00 	imes 10^{-2} \ g}{30 \ g \ mol^{-1}}$.Using the proportionality $p_2 / p_1 = n_2 / n_1$:$p_2 = p_1 	imes (n_2 / n_1) = 1 \ bar 	imes (5.00 	imes 10^{-2} / 6.56 	imes 10^{-2})$.$p_2 = 5.00 / 6.56 \ bar = 0.762 \ bar \approx 0.764 \ bar$ (considering standard rounding).Therefore,the partial pressure is $0.764 \ bar$.

The partial pressure of ethane over a solution containing $6.56 \times 10^{-2} \ g$ of ethane is $1 \ bar$. If the solution contains $5.00 \times 10^{-2} \ g$ of ethane,then what shall be the partial pressure of the gas (in $bar$)?

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