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

Question

Henry's law constant for the molality of methane in benzene at 298 K is 4.27× 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg. (A)  1.78× 10-4  (B)  1.78× 10-3  (C)  1.78× 104  (D)  1.78× 10-6

Tardigrade Admin · Accepted Answer

K H =4.27 × 105 mm Hg ( at 298 K ) p =760 mm Applying Henry's law, p = K H x , where x = Mole fraction or solubility of methane. x =( p / K H )=((760 mm )/(4.27 × 105 mm )) =1.78 × 10-3

Q. Henry's law constant for the molality of methane in benzene at $298 K$ is $4.27 \times 1 0^{5} mm H g$ . Calculate the solubility of methane in benzene at $298 K$ under $760 mm H g$ .

$1.78 \times 1 0^{- 4}$

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

$1.78 \times 1 0^{4}$

$1.78 \times 1 0^{- 6}$

$K_{H} = 4.27 \times 1 0^{5} mm H g ($ at $298 K)$
$p = 760 mm$
Applying Henry's law,
$p = K_{H x},$ where $x =$ Mole fraction or solubility of methane.
$x = \frac{p}{K _{H}} = \frac{( 760 mm )}{( 4.27 \times 1 0 ^{5} mm )}$
$= 1.78 \times 1 0^{- 3}$

Q. Henry's law constant for the molality of methane in benzene at 298K is 4.27×105mmHg. Calculate the solubility of methane in benzene at 298K under 760mmHg.

1.78×10−4

1.78×10−3

1.78×104

1.78×10−6