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Q.

A company dissolves 'X' amount of CO₂at 298 K in 1 litre of water to prepare soda water $X =______\times 10^{- 3} g .$ (nearest integer)
(Given: partial pressure of ${CO}_{2} at 298 K = 0.835$
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)

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answer is 1222.

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Detailed Solution

From Henry law

$\begin{aligned} P = K_{H} X_{{CO}_{2}} \\ 0 .835 = 1 .67 \times 10^{3} \times 1 .67 \times 10^{3} \times \frac{W_{{CO}_{2}} / 44}{\frac{W_{{CO}_{2}}}{44} + \frac{1000}{18}} \\ w_{{CO}_{2}} = 1 .2228 g = 1222 .8 \times 10^{- 3} g \end{aligned} Or \begin{aligned} P = K_{H} X_{{CO}_{2}} \\ 0 .835 = 1 .67 \times 10^{3} \times \frac{n_{{CO}_{2}}}{n_{{CO}_{2}} + n_{H_{2} O}} \\ 0 .835 = 1 .67 \times 10^{3} \times \frac{w_{{CO}_{2}} / 44}{\frac{1000}{18}} \\ w_{{CO}_{2}} = 1 .2222 g = 1222 .2 \times 10^{- 3} g \end{aligned}$

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A company dissolves 'X' amount of CO2 at 298 K in 1 litre of water to prepare soda water X=______×10-3 g. (nearest integer) (Given: partial pressure of CO2 at 298K=0.835 Henry's law constant for CO2 at 298 K=1.67 kbar.Atomic mass of H,C and O is 1,12 and 6 g mol-1, respectively)