# Question 5b7ab

May 4, 2016

$4 \cdot {10}^{- 4} {\text{moles N}}_{2}$

#### Explanation:

For starters, I think that you mistyped the value of the Henry law constant, it should actually be

${K}_{H} = 1 \cdot {10}^{5} \text{atm" " }$ or ${\text{ " K_H = 1 * 10^(-5)"atm}}^{- 1}$

I will use ${K}_{H} = 1 \cdot {10}^{5} \text{atm}$, since I assume that this was the actual value given to you.

The idea here is that you can use Dalton's Law of Partial Pressures to determine the partial pressure of nitrogen gas, ${\text{N}}_{2}$, above the solution, then use Henry' Law to determine the mole fraction of nitrogen gas.

As you know, Henry' Law states that the solubility of a gas in a liquid is proportional to the partial pressure of the gas above the liquid.

If you take ${\chi}_{{N}_{2}}$ to be the mole fraction of nitrogen gas in the solution, you can express Henry's Law as

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{b l a c k}{\textcolor{w h i t e}{\frac{a}{a}} P = {K}_{H} \times {\chi}_{{N}_{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that the partial pressure of nitrogen gas above the solution depends on its mole fraction in the solution.

Now, the partial pressure of nitrogen gas above the solution will depend on its mole fraction in the air $\to$ think Dalton's Law of Partial Pressures here.

You will have

color(purple)(|bar(ul(color(white)(a/a)color(black)(P_(N_2) = chi_("air N"_2) * P_"total")color(white)(a/a)|)))#

Here

${P}_{{N}_{2}}$ - the partial pressure of nitrogen gas
${P}_{\text{total}}$ - the total pressure above the solution
${\chi}_{{\text{air N}}_{2}}$ - the mole fraction of ${\text{N}}_{2}$ in the air, equal to $0.8$

Plug in your values to get

${P}_{{N}_{2}} = 0.8 \cdot \text{5 atm" = "4 atm}$

Plug this into the equation for Henry's Law to get the mole fraction of nitrogen gas dissolved in the solution

$4 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) = 1 * 10^(5)color(red)(cancel(color(black)("atm}}}} \cdot {\chi}_{{N}_{2}}$

${\chi}_{{N}_{2}} = \frac{4}{1 \cdot {10}^{5}} = 4 \cdot {10}^{- 5}$

SIDE NOTE This is why I said that the value you listed for the Henry law constant is inaccurate. Assuming that you have ${K}_{H} = 1 \cdot {10}^{- 5} \text{atm}$, the mole fraction of dissolved nitrogen gas in this solution will come out to be $> 1$, which is not possible

${\chi}_{{N}_{2}} = \frac{4}{1 \cdot {10}^{- 5}} = 4 \cdot {10}^{5} \to$ NOT possible

Now, you know the mole fraction of dissolved nitrogen gas in this solution. This mole fraction tells you the ratio that exists between the number of moles of nitrogen gas and the total number of moles present in the solution.

If you take $n$ to be the number of moles of nitrogen gas dissolved, you can say that

${\chi}_{{N}_{2}} = \frac{n}{n + 10}$

This is true because you know that you have $10$ moles of water present in this solution.

You thus have

$4 \cdot {10}^{- 5} = \frac{n}{n + 10}$

Rearrange to get

$n = 4 \cdot {10}^{- 5} \cdot n + 4 \cdot {10}^{- 4}$

$n \left(1 - 4 \cdot {10}^{- 5}\right) = 4 \cdot {10}^{- 4}$

This will be equivalent to

$n = \frac{4 \cdot {10}^{- 4}}{1 - 4 \cdot {10}^{- 5}} \approx \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 4 \cdot {10}^{- 4} {\text{moles N}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$