Henry's law constant for 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 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) 2.56× 10- 4 (B) 1.78× 10- 3 (C) 3.78× 10- 2 (D) 4.13× 10- 1

Tardigrade Admin · Accepted Answer

KH = 4.27 &times; 105 mm Hg (at 298 K) p = 760 mm Applying Henry's law p = KH x [x = Mole fraction / solubility of methane] textx = ( textp/( textK) textH) = (( text760 mm)/( text4.27 × ( text10)5 textmm)) = text178 × ( text10)- 5 = text1.78 × ( text10)- 3

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

$2.56 \times 1 0^{- 4}$

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

$3.78 \times 1 0^{- 2}$

$4.13 \times 1 0^{- 1}$

K_H = 4.27 × 10⁵ mm Hg (at 298 K)
p = 760 mm
Applying Henry's law
p = K_H x [x = Mole fraction / solubility of methane]
$x = \frac{p}{( K ) _{H}} = \frac{( 760 mm )}{( 4.27 \times ( 10 ) ^{5} mm )} = 178 \times (10)^{- 5} = 1.78 \times (10)^{- 3}$

Q. Henry's law constant for 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.

2.56×10−4

1.78×10−3

3.78×10−2

4.13×10−1