At 310 K the partial vapor pressures of a substance B dissol

At 310 K, the partial vapor pressures of a substance B dissolved in a liquid A are as follows:

xB: 0.010, 0.015, 0.020

pB: 82.0, 122.0, 166.1

Show that the solution obeys Henry\'s law in this range of mole fractions, and calculate Henry\'s law constant at 310 K

Solution

Here

According to the Henry Law

Partial pressure = henry Constant*mole Fraction

Therefore

p = kx

82 = 0.010*k

k = 8200 atm/mol

The ratio (p/x) approx remains Constant

This gives the reault that Henry Law is Valid

From Second

k = 122/0.015

= 8133.33 atm/mole

For Third

k = 166.1/0.020

= 8305 atm/mole

Thereofre

Kreqd = Average of three values of k

= (8200+8133.33+8305)/3

= 8212.78 atm/mole