Question

The Henry’s law constant for the solubility of ${{\text{N}}_{2}}$ gas in water at 298K is 1.0×${{10}^{5}}$atm. The mole fraction of ${{\text{N}}_{2}}$ in air is 0.8. the number of moles of ${{\text{N}}_{2}}$ from air dissolved in 10 moles of water at 298K and 5 atm pressure is:
(a) 4.0×${{10}^{-4}}$
(b) 4.0×${{10}^{-5}}$
(c) 5.0×${{10}^{-4}}$
(d) 4.0×${{10}^{-6}}$

Answer 1

First we have to find the partial pressure of the gas ${{\text{N}}_{2}}$ from the total pressure and the mole fraction given and then by using the Henry’s formula of p=${{K}_{H}}$ x, we can find the moles of the ${{\text{N}}_{2}}$ gas. Here, p is the partial pressure of the gas ${{\text{N}}_{2}}$, ${{K}_{H}}$ is the Henry’s constant and x is the mole fraction of ${{\text{N}}_{2}}$ and water. Now, solve it.

Complete step by step answer:
First of all, we should know first what Henry's law is. Henry’s law states that if the solution of the gas in the liquid , the gaseous component is volatile then, its solubility is given as :
p=${{K}_{H}}$ x ---------(A)
here, p is the partial pressure of the gas in the solution and x is the mole fraction and ${{K}_{H}}$ is called as the Henry’s constant and is equal to the vapour pressure of the pure component $p_{A}^{\circ }$ .
Now, considering the numerical; first we have to find the partial pressure of the gas ${{\text{N}}_{2}}$(p) by applying the formula as:
p=P × x ----------(B)
As we know that the total pressure of the gas(P)= 5 atm (given)
And mole fraction of ${{\text{N}}_{2}}$(x)=0.8
Put all these values in equation(B),we get;
p = 5 × 0.8
p =4 atm -----------(1)
So, the partial pressure of the gas is 4atm.

Now, we will calculate the mole fraction of nitrogen gas from the mixture of nitrogen and water by the formula as:
Mole fraction =$\dfrac{\text{no of moles of solute }(y)}{\text{no of moles of mixture}}$
As we, know that no of moles of mixture (water+nitrogen)=10 (given)
Let suppose no of moles of the solute (nitrogen gas)=y
Then, mole fraction is ;
Mole fraction =$\dfrac{y}{10}$ --------(2)
Henry’s constant =1.0×${{10}^{5}}$ atm -------(3)

Now, put all these values of equation (1),(2) and (3) in equation(A), we get;
4 =1.0×${{10}^{5}}$× $\dfrac{y}{10}$
$\dfrac{4\times 10}{1\times {{10}^{5}}}$ =y
y= $4\times {{10}^{-4}}$ mol
so, thus no of moles of nitrogen gas is $4\times {{10}^{-4}}$ mol
So, the correct answer is “Option A”.

Note: This law is not applicable at low pressure and temperature because at the it, the law becomes less accurate and the proportionality constant ${{K}_{H}}$ shows considerable deviations and the gas in this neither reacts chemically nor undergoes any association or dissociation in the solvent.