The partial pressure of ethane over a solutio | vedclass

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(C) According to Henry's law,the partial pressure of a gas $(p)$ is directly proportional to its mass $(m)$ dissolved in a given volume of solvent at

Vedclass · Accepted Answer

$7$

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(C) According to Henry's law,the partial pressure of a gas $(p)$ is directly proportional to its mass $(m)$ dissolved in a given volume of solvent at a constant temperature: $p \propto m$ or $p = k 	imes m$. Given: $p_1 = 1 \ bar$,$m_1 = 7.14 	imes 10^{-3} \ g$. We need to find $p_2$ when $m_2 = 5 	imes 10^{-2} \ g$. Using the ratio: $\frac{p_2}{p_1} = \frac{m_2}{m_1}$. $\frac{p_2}{1} = \frac{5 	imes 10^{-2}}{7.14 	imes 10^{-3}} = \frac{50 	imes 10^{-3}}{7.14 	imes 10^{-3}} \approx 7 \ bar$. Thus,the partial pressure is $7 \ bar$.

The partial pressure of ethane over a solution containing $7.14 \times 10^{-3} \ g$ of ethane is $1 \ bar$. If the solution contains $5 \times 10^{-2} \ g$ of ethane,then what shall be the partial pressure of the gas (in $bar$)? (At equal temperature)

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