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Chemistry Question on Solutions

Henry’s law constant for the molality of methane in benzene at 298 K is 4.27×105 mm Hg4.27 \times 105\, mm\, Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Answer

The correct answer is: 178×10−5 178 \times 10^{ - 5}
Here,
p = 760 mm Hg
kH=4.27×105mmHgk_H = 4.27 \times 10^ 5 mm Hg
According to Henry's law,
p=kHxp = k_Hx
⇒x=pKH⇒x=\frac{p}{K_H}
=760mmHg4.27×105mmHg=\frac{760mm Hg}{4.27\times10^5mmHg}
=177.99×10−5= 177.99 \times 10^{ - 5 }
=178×10−5= 178 \times 10^{ - 5} (approximately)
Hence, the mole fraction of methane in benzene is 178×10−5 178 \times 10^{ - 5}