Question

9) The mole fraction of gas dissolved in a solvent is given by Henry's law. If the Henry's law
constant for a gas in water at 298 K is 5.55 x 10? Torr and the partial pressure of the gas is 200
Torr, then what is the amount of the gas dissolved in 1.0 kg of water?
a) 2.0 x 10-4 mol b) 2.5 x 10-5 mol c) 3.7 x 10-6 mol d) 1.2 x 10-8 mol​

Answer 1

Answer:

Henry's law constant for the solubility of nitrogen gas in water at

is

. The mole fraction of nitrogen in air is

.The number of moles of nitrogen from air dissolved in

of water at

and

pressure is

Answer 2

Given:

According to the question we have

${p_{A}} =200 \mathrm{Torr}\\\\{K_{H}}=5.55 \times 10^{7} \mathrm{Torr}}$

So By Henry's law

$p_{A}=K_{H} \times x_{A}$

$x_{A}=\frac{p_{A}}{K_{H}}\\\\=\frac{200 \mathrm{Torr}}{5.55 \times 10^{7} \mathrm{Torr}}\\\\=3.6 \times 10^{-6}$

$\text { But } x_{A}=\frac{n_{A}}{n_{A}+n_{H_{2} O}} \cong \frac{n_{A}}{n_{H_{2} O}}\\\\x_{A}=\frac{n_{A}}{1000 / 18}$

$n_{A}=m_{A} \times \frac{1000}{18}\\\\=3.6 \times 10^{-6} \times \frac{1000}{18} \text { mole }\\\\=2.0 \times 10^{-4} \mathrm{~mole}$

The correct option is "a"

Hence the answer is $2.0 \times 10^{-4} \mathrm{~mole}$