A company dissolves '$$x$$' amount of $$CO_2$$ at $$298$$ K in $$1$$ litre of water to prepare soda water. $$X = $$ ______ $$\times 10^{-3}$$ g. (nearest integer)
(Given: partial pressure of $$CO_2$$ at $$298$$ K $$= 0.835$$ bar. Henry's law constant for $$CO_2$$ at $$298$$ K $$= 1.67$$ kbar. Atomic mass of H, C and O is $$1, 12,$$ and $$6$$ g mol$$^{-1}$$, respectively)

We need to find the mass of $$CO_2$$ dissolved in 1 litre of water at 298 K.

Since Henry’s Law relates the partial pressure of a gas to its mole fraction in solution, we write

$$ p_{CO_2} = K_H \cdot x_{CO_2} $$

Substituting the given values gives

$$ x_{CO_2} = \frac{p_{CO_2}}{K_H} = \frac{0.835 \text{ bar}}{1.67 \text{ kbar}} = \frac{0.835}{1670} = 5 \times 10^{-4} $$

Next, the number of moles of water in 1 litre is

$$ n_{H_2O} = \frac{1000}{18} = 55.56 \text{ mol} $$

Since $$x_{CO_2}$$ is very small, we approximate

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

This gives

$$ n_{CO_2} = x_{CO_2} \times n_{H_2O} = 5 \times 10^{-4} \times 55.56 = 0.02778 \text{ mol} $$

Calculating the mass of $$CO_2$$ (molar mass of $$CO_2$$ = 12 + 2(16) = 44 g/mol) yields

$$ m_{CO_2} = 0.02778 \times 44 = 1.2222 \text{ g} $$

Expressing this value in the required form results in

$$ x = 1.2222 \text{ g} \approx 1221 \times 10^{-3} \text{ g} $$

The answer is $$1221$$.