Henry-s law constant for the molality of methane in benzene at 298 Kis 4-27 - 105 mm Hg - Calculate the solubility of methane in benzene at 298 Kunder760 mm Hg-

Here, p = 760 mm Hg kH = 4.27 × 105 mm Hg According to Henry’s law, p = kHx = 177.99 × 10−5 = 178 × 10−5 (approximately) Hence, the mole fraction of met ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Relative Lowering of Vapour Pressure

Henry's law c...

Question

Henry’s law constant for the molality of methane in benzene at 298 Kis 4.27 × 10⁵ mm Hg. Calculate the solubility of methane in benzene at 298 Kunder 760 mm Hg.

Open in App

Solution

Here,

p = 760 mm Hg

k_H= 4.27 × 10⁵ mm Hg

According to Henry’s law,

p = k_Hx

= 177.99 × 10⁻⁵

= 178 × 10⁻⁵ (approximately)

Hence, the mole fraction of methane in benzene is 178 × 10⁻⁵.

Suggest Corrections

Similar questions

Q. Henry's law constant for the molality of methane in benzene at 298 K is

4.27 \times 10^{5}

H g

. Calculate the solubility of methane in benzene at 298 K under 760 mm

H g

Q. Henry's law constant for the molality of methane in benzene at

298 K

4.27 \times 105 m m H g

. Calculate the solubility of methane in benzene at

298 K

under

760 m m H g

Q. Henry's law constant for molality of methane in benzene at

298 K

4.27 \times 10^{5} m m

H g

. The mole fraction of methane in benzene at

298 K

under

760 m m

H g

is:

Q. Henry's law constant for solubility of methane in benzene is

4.27 \times 10^{- 5}

m m^{- 1} H g m o l

d m^{- 3}

at constant temperature. Calculate solubility of methane at

760 m m H g

pressure at same temperature.

Q. Henry law constant for the solubility of methane in benzene at 298K is

4.27 \times 10^{5}

mm Hg then the solubility of methane in benzene at 298K under 760mm Hg is:

Join BYJU'S Learning Program

Henry’s law constant for the molality of methane in benzene at 298 Kis 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 Kunder 760 mm Hg.

Here, p = 760 mm Hg kH = 4.27 × 105 mm Hg According to Henry’s law, p = kHx = 177.99 × 10−5 = 178 × 10−5 (approximately) Hence, the mole fraction of methane in benzene is 178 × 10−5.

Henry’s law constant for the molality of methane in benzene at 298 Kis 4.27 × 10⁵ mm Hg. Calculate the solubility of methane in benzene at 298 Kunder 760 mm Hg.

Here,

p = 760 mm Hg

k_H= 4.27 × 10⁵ mm Hg

According to Henry’s law,

p = k_Hx

= 177.99 × 10⁻⁵

= 178 × 10⁻⁵ (approximately)

Hence, the mole fraction of methane in benzene is 178 × 10⁻⁵.