Question

The Henry's law constant at ${\mathbf{25}}^{\mathbf{0}}\mathit{C}$for carbon dioxide dissolved in water is$\mathbf{1}\mathbf{.}\mathbf{65}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{\text{atm}}$ . If a carbonated beverage is bottled under a ${\mathbf{\text{CO}}}_{\mathbf{2}}$pressure of 5.0 atm .

(a) Calculate the number of moles of carbon dioxide dissolved per liter of water under these conditions, using$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{g}}\mathbf{}{\mathbf{\text{cm}}}^{\mathbf{-}\mathbf{3}}$ as the density of water.

(b) Explain what happens on a microscopic level after the bottle cap is removed.

Short answer

(a) $n_{{\text{CO}}_2}=0.17\text{mol}$

(b) The partial pressure of ${\text{CO}}_2$in the atmosphere is less than 1 atm, the excess ${\text{CO}}_2$will bubble out from the solution when it opens.

Step-by-step solution

Given data

The molar mass of the polymer $7.4\times {10}^3{\text{gmol}}^{-1}$, ${\text{CO}}_2$ pressure is $5.0atm$, density of water is $1.00\text{g}{\text{cm}}^{-3}$.

Concept used

To determine the mole fraction of a sample, apply Henry's law. According to the law, the partial pressure of a gas in equilibrium with a given liquid is directly proportional to the amount of that gas that dissolves in the liquid at a given temperature.

Calculate the mole fraction

Henry’s law is used to calculate the mole fraction of carbon dioxide in water;

$X_{{\text{CO}}_2}=\frac{P_{{\text{CO}}_2}}{k_{{\text{CO}}_2}}$

$$\begin{gathered} X_{{\text{CO}}_2}=\frac{5.0\text{atm}}{1.65\times {10}^3\text{atm}} \\ =3.03\times {10}^{-3} \end{gathered}$$

The mole fraction is converted to molarity. One liter of water weighs 1000g, so it contains

$n_{{\text{H}}_2\text{O}}=\frac{1000\text{g}}{18.02{\text{gmol}}^{-1}}=55.5\text{mol}$water

Calculate the moles of  

Because$X_{{\text{CO}}_2}$is too small,$n_{{\text{H}}_2\text{O}}+n_{{\text{CO}}_2}$is close to$n_{{\text{H}}_2\text{O}}$, and it can be written as:

$X_{{\text{CO}}_2}=\frac{n_{{\text{CO}}_2}}{n_{{\text{H}}_2\text{O}}+n_{{\text{CO}}_2}}\approx \frac{n_{{\text{CO}}_2}}{n_{{\text{H}}_2\text{O}}}$

The number of moles of ${\text{CO}}_2$

$$\begin{gathered} n_{{\text{CO}}_2}=X_{{\text{CO}}_2}\times n_{{\text{H}}_2\text{O}} \\ =3.03\times {10}^{-3}\times 55.5\text{mol} \\ =0.17\text{mol} \end{gathered}$$

The microscopic result when bottle cap is removed

The partial pressure of ${\text{CO}}_2$ in the atmosphere is less than 1 atm , the excess will bubble out from the solution when it opens.