Question

At 298 K, 500cm3 H2O dissolved 15.30
cm3 CHA(STP) under a partial pressure
of methane of one atm. If Henery's law
holds, what pressure is required to
cause 0.001 mole methane to dissolve
in 300cmwater?
(a) 0.286 atm
(b) 2.486 atm
(c) 1.286 atm
(d) 3.111 atm
1​

Answer 1

Given

R = Gas constant = $82.05\ \text{cm}^3\text{atm/mol K}$

T = Standard temperature = 273 K

P = Standard pressure = 1 atm

M = Molar mass of methane = 16 g/mol

p = Partial pressure of methane = 1 atm

k = Henry's constant

$500\ \text{cm^3}$ of $H_2O$ dissolved in $15.30\ \text{cm}^3$ of $CH_4$

To find

Pressure is required to cause 0.001 mole methane to dissolve in 300 cm³ of water.

Solution

Volume of $CH_4$  dissolved in $1\ \text{cm}^3$ of water is

$\dfrac{15.3}{500}=0.0306\ \text{cm}^3$

From the ideal gas law we get

$n=\dfrac{PV}{RT}\\\Rightarrow n=\dfrac{1\times 0.0306}{82.05\times 273}\\\Rightarrow n=1.36\times 10^{-6}\ \text{moles}$

Mass of methane

$m=nM\\\Rightarrow m=1.36\times 10^{-6}\times 16\\\Rightarrow m=2.176\times 10^{-5}\ \text{g}$

From Henry's law we have

$m=kp\\\Rightarrow k=\dfrac{m}{p}\\\Rightarrow k=\dfrac{2.176\times 10^{-5}}{1}\\\Rightarrow k=2.176\times 10^{-5}\ \text{g/atm}$

For the 0.001 mole of methane dissolved in 1 cc of water we have

$\dfrac{0.001\times 16}{300}=5.33\times 10^{-5}\ \text{g}$

From Henry's constant (k) we have

$p=\dfrac{m}{k}\\\Rightarrow p=\dfrac{5.33\times 10^{-5}}{2.176\times 10^{-5}}\\\Rightarrow p=2.44944\ \text{atm}\approx 2.486\ \text{atm}$

The required pressure is (b) 2.486 atm