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 m mol 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)

We start with Henry’s law, which states that for a gas dissolved in a liquid at constant temperature, the partial pressure of the gas in the vapour phase is directly proportional to its mole fraction in the solution:

$$p = k_H \, x_{\text{CO}_2}$$

Here, $$p = 0.835 \text{ bar}$$ is the partial pressure of CO$$_2$$, and $$k_H = 1.67 \times 10^{3} \text{ bar}$$ is the Henry’s law constant for CO$$_2$$ at 298 K. Rearranging the formula to find the mole fraction $$x_{\text{CO}_2}$$, we have:

$$x_{\text{CO}_2} = \frac{p}{k_H}$$

Substituting the given values,

$$x_{\text{CO}_2} = \frac{0.835 \text{ bar}}{1.67 \times 10^{3} \text{ bar}}$$

$$x_{\text{CO}_2} = \frac{0.835}{1670}$$

$$x_{\text{CO}_2} = 5.0 \times 10^{-4}$$

Now we let $$n_{\text{CO}_2}$$ be the number of moles of CO$$_2$$ that dissolve in the water and $$n_{\text{H}_2\text{O}}$$ be the moles of water present. The mole fraction definition gives:

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

We next determine $$n_{\text{H}_2\text{O}}$$. The volume of water is 0.9 L. Assuming the density of water is $$1.0 \text{ g mL}^{-1}$$ (i.e., $$1.0 \text{ g cm}^{-3}$$), the mass of 0.9 L (which is $$900 \text{ mL}$$) is:

$$m_{\text{H}_2\text{O}} = 900 \text{ g}$$

The molar mass of water is $$18 \text{ g mol}^{-1}$$, so:

$$n_{\text{H}_2\text{O}} = \frac{m_{\text{H}_2\text{O}}}{M_{\text{H}_2\text{O}}} = \frac{900 \text{ g}}{18 \text{ g mol}^{-1}} = 50 \text{ mol}$$

Because we expect $$n_{\text{CO}_2}$$ to be very small compared to $$n_{\text{H}_2\text{O}}$$, we can approximate $$n_{\text{CO}_2} + n_{\text{H}_2\text{O}} \approx n_{\text{H}_2\text{O}}$$. Thus,

$$x_{\text{CO}_2} \approx \frac{n_{\text{CO}_2}}{n_{\text{H}_2\text{O}}}$$

Substituting the known values,

$$5.0 \times 10^{-4} \approx \frac{n_{\text{CO}_2}}{50}$$

Multiplying both sides by 50,

$$n_{\text{CO}_2} \approx 50 \times 5.0 \times 10^{-4}$$

$$n_{\text{CO}_2} \approx 2.5 \times 10^{-2} \text{ mol}$$

Converting moles to millimoles (1 mol = 1000 mmol),

$$n_{\text{CO}_2} \approx 2.5 \times 10^{-2} \text{ mol} \times 1000 \frac{\text{mmol}}{\text{mol}} = 25 \text{ mmol}$$

Therefore, $$x = 25$$.

So, the answer is $$25$$.