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A company dissolves 'x' amount of $\mathrm{CO_{2}}$ at 298 K in 1 litre of water to prepare soda water. $\mathrm{ X=}$ ____________ $\mathrm{ \times10^{-3}}$ . (nearest integer)

(Given: partial pressure of $\mathrm{CO_{2}}$ , at 298 K = 0.835 bar.

Henry's law constant for $\mathrm{CO_{2}}$ at 298K = 1.67 kbar.

Atomic mass of H, C and O is 1, 12, and 16 g $\mathrm{mol^{-1}}$ , respectively)
Option: 1
1221

Option: 2
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Option: 3
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Option: 4
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Answers (1)

best_answer

We have to find amount of $\mathrm{CO}_{2}$ dissolved, in 1 litre of $\mathrm{H}_{2} \mathrm{O}$

Given,

$\mathrm{P_{CO_2}=0.835\ bar}$ ,

Henry's law constant, $\mathrm{K_H=1.67\ kbar}$

Using Henry's Law, we have

$\mathrm{P=K_{H} X, \quad X \text { is mole fraction }}$

Let the mass of $\mathrm{CO_2}$ dissolved be $x$ . So we can write

$0.835=1.67 \times 10^{3} \times\left(\frac{\frac{x}{M_{CO_{2}}}}{\frac{x}{M_{\mathrm{CO}_{2}}}+\frac{1000}{18}}\right) \quad\left(\because \mathrm{lml=1 g} \; \text{of}\; \mathrm{H_{2} O}\right)$

$0.835=1.67 \times 10^{3}\left(\frac{\frac{x}{44}}{\frac{x}{44}+55.5}\right)=1.67 \times 10^{3}\left(\frac{\frac{x}{44}}{55.5}\right)$

Now, assuming a very dilute solution, we can approximate as

$55.5>> \frac{x}{44} \Rightarrow 55.5+\frac{x}{44} \simeq 55.5$

Then,

$\frac{x}{44}=\frac{0.835 \times 55.5}{1.67 \times 10^{3}} \Rightarrow x \approx 1221 \times 10^{-3} \mathrm{~g}$

Correct answer is 1221

Posted by

Sanket Gandhi

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