Henry law constant for the solubility of methane in benzene at 298K is 4.27×10⁵ mm Hg then the solubility of methane in benzene at 298K under 760mm Hg is

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7.8 × 10^-2 mole/kg

7.8 × 10^-4mole/kg

7.8 × 10^-3mole/kg

1.78 × 10^-3mole/kg

answer is D.

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Detailed Solution

Given Data:

K_H = 4.27 × 10⁵ mm Hg
p = 760 mm Hg

Explanation

Rearranging the formula to solve for x:

x = p / K_H = 760 / (4.27 × 10⁵) ≈ 1.78 × 10^-3

Therefore, the mole fraction of methane in benzene is approximately 1.78 × 10^-3.

Expressing Solubility as Molality

To express this solubility in terms of molality (moles of solute per kilogram of solvent), we use the approximation that for dilute solutions, the mole fraction is roughly equal to the molality when the solvent molar mass is large compared to the solute.

Since benzene has a molar mass of about 78 g/mol, this approximation holds reasonably well.

Final Answer

Thus, the solubility of methane in benzene at 298 K under 760 mm Hg is approximately 1.78 × 10^-3 mole/kg, corresponding to option d.

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