Henry's law constant for the molality of methane in benzene at 298 K is 4.27 x 10⁵mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Solution

Answer:

p = 760mmHg
K_H = 4.27 x 10⁵

According to Henry's law
v = K_Hx
k_H = 4.27 x 10⁵ mm Hg,
p = 760 mm

760 = 4.27 \times 10^{5} \times x

x = \frac{760}{4.27 \times 10^{5}} = 178 \times 10^{- 5} .

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