Henry's law constant for the dissolution of in benzene at is . What is the solubility of in benzene — Solutions and Colligative Properties Chemistry Question
Question
Henry's law constant for the dissolution of $CH_4$ in benzene at $298\text{ K}$ is $2 \times 10^5\text{ mm of Hg}$. What is the solubility of $CH_4$ in benzene at $298\text{ K}$ (expressed as mole fraction) under an equilibrium pressure of $760\text{ mm of Hg}$?
Answer: B
💡 Solution & Explanation
According to Henry's law, $P = K_H \times \chi$. Given $P = 760\text{ mm Hg}$ and $K_H = 2 \times 10^5\text{ mm Hg}$. The mole fraction $\chi = \frac{P}{K_H} = \frac{760}{2 \times 10^5} = 3.8 \times 10^{-3}$.
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