Question

$CO_2$ gas is bubbled through water during a soft drink manufacturing process at 298 K. If $CO_2$ exerts a partial pressure of 0.835 bar, then x mmol of $CO_2$ would dissolve in 0.9 L of water. The value of x is ______. (Nearest integer) (Henry's law constant for $CO_2$ at 298 K is 1.67 $\times 10^3$ bar)

Answer 1

25

Answer 2

According to Henry's Law, the partial pressure of a gas ($P_{gas}$) above a liquid is proportional to the mole fraction of the gas in the solution ($X_{gas}$):

$P_{gas} = K_H \cdot X_{gas}$

Given:

• Partial pressure of $CO_2$, $P_{CO_2} = 0.835$ bar

• Henry's law constant for $CO_2$, $K_H = 1.67 \times 10^3$ bar

We can calculate the mole fraction of $CO_2$ in the water:

$X_{CO_2} = \frac{P_{CO_2}}{K_H} = \frac{0.835 \text{ bar}}{1.67 \times 10^3 \text{ bar}} = \frac{0.835}{1670} = 0.0005 = 5 \times 10^{-4}$

The mole fraction of $CO_2$ is defined as:

$X_{CO_2} = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}}$

where $n_{CO_2}$ is the number of moles of dissolved $CO_2$ and $n_{H_2O}$ is the number of moles of water.

The volume of water is 0.9 L. Assuming the density of water is 1 g/mL (or 1000 g/L), the mass of water is:

Mass of water = Volume $\times$ Density = 0.9 L $\times$ 1000 g/L = 900 g

The molar mass of water ($H_2O$) is approximately 18 g/mol (using atomic masses H=1, O=16).

The number of moles of water is:

$n_{H_2O} = \frac{\text{Mass of water}}{\text{Molar mass of water}} = \frac{900 \text{ g}}{18 \text{ g/mol}} = 50 \text{ mol}$

Since the solubility of gases in liquids is generally low, the number of moles of dissolved $CO_2$ ($n_{CO_2}$) is much smaller than the number of moles of water ($n_{H_2O}$). Therefore, we can approximate $n_{CO_2} + n_{H_2O} \approx n_{H_2O}$.

The mole fraction equation becomes:

$X_{CO_2} \approx \frac{n_{CO_2}}{n_{H_2O}}$

Now we can calculate the number of moles of $CO_2$:

$n_{CO_2} = X_{CO_2} \times n_{H_2O}$

$n_{CO_2} = (5 \times 10^{-4}) \times 50 \text{ mol}$

$n_{CO_2} = 250 \times 10^{-4} \text{ mol} = 0.025 \text{ mol}$

The question asks for the amount of $CO_2$ dissolved in millimoles (x mmol).

$x \text{ mmol} = n_{CO_2} \times 1000 \text{ mmol/mol}$

$x = 0.025 \text{ mol} \times 1000 \text{ mmol/mol}$

$x = 25 \text{ mmol}$

The value of x is 25. We need to provide the nearest integer, which is 25.