Henry’s law constant for molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. The mole fraction of methane in benzene at 298 K under 760 mm Hg is

Question

Henry’s law constant for molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. The mole fraction of methane in benzene at 298 K under 760 mm Hg is (A) 1.78 × 10-3 (B) 17.43 (C) 0.114 (D) 2.814

Tardigrade Admin · Accepted Answer

According to Henryâs law, p = KHx x = (p/KH) = (760/4.27 × 105) = 1.78 × 10-3

Q. Henry’s law constant for molality of methane in benzene at $298 K$ is $4.27 \times 1 0^{5} mm H g$ . The mole fraction of methane in benzene at $298 K$ under $760 mm H g$ is

$1.78 \times 1 0^{- 3}$

$17.43$

$0.114$

$2.814$

According to Henry’s law, $p = K_{H} x$
$x = \frac{p}{K _{H}} = \frac{760}{4.27 \times 1 0 ^{5}}$
$= 1.78 \times 1 0^{- 3}$

Q. Henry’s law constant for molality of methane in benzene at 298K is 4.27×105mmHg. The mole fraction of methane in benzene at 298K under 760mmHg is

1.78×10−3

17.43

0.114

2.814