Use Henry’s law

For a gas dissolved in a liquid, $p = K_H x$ where:

• $p = 0.835\,\text{bar}$
• $K_H = 1.67\,\text{kbar} = 1670\,\text{bar}$
• $x$ = mole fraction of dissolved $CO_2$

So, $x = \frac{p}{K_H} = \frac{0.835}{1670} = 5.0\times 10^{-4}$

Relate mole fraction to moles dissolved

Let moles of dissolved $CO_2$ be $n$.

In $1$ litre water, mass of water $\approx 1000\,\text{g}$.

Molar mass of water: $M(H_2O)=2(1)+16=18\,\text{g mol}^{-1}$

Hence moles of water, $n_{H_2O} = \frac{1000}{18} = 55.56$

Mole fraction of $CO_2$ is $x = \frac{n}{n+55.56}$

Since $x=5.0\times 10^{-4}$ is very small, we may use $x \approx \frac{n}{55.56}$

Therefore, $n = x\times 55.56 = 5.0\times 10^{-4}\times 55.56 = 2.778\times 10^{-2}\,\text{mol}$

Convert moles of $CO_2$ to mass

Molar mass of $CO_2$: $M(CO_2)=12+2(16)=44\,\text{g mol}^{-1}$

Mass dissolved, $m = n\times 44 = 2.778\times 10^{-2}\times 44 = 1.222\,\text{g}$

Express in the required form

Given: $X = \underline{\hspace{2cm}}\times 10^{-3}\,\text{g}$

Since $1.222\,\text{g} = 1222\times 10^{-3}\,\text{g}$

So the nearest integer is $\boxed{1222}$

Check against stored answer

Stored correct answer = $1221$.

Our calculated value is about $1222$. Even using the exact expression, $n = \frac{x\,n_{H_2O}}{1-x}$ gives essentially the same result, still nearest integer $1222$.

Hence the stored answer appears off by $1$, likely due to rounding/truncation.