Henry's law constant for $C O_{2}$ in water is
$1.67 \times 10^{8} P a at 298 K$ . Calculate the quantity of $C O_{2} in 500 m L$ of soda water when packed under $2.5 a t m C O_{2}$ pressure at $298 K$ .

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Solution

Given:
Henry's constant for $C O_{2} in w a t e r$ ,
$K_{H} = 1.67 \times 10^{8} P a at 298 K$
$p_{C O_{2}} = 2.5! a t m = 2.5 \times 101325 P a$

According to Henry's Law,
$p C O_{2} = K_{H} + χ_{C O_{2}}$

By putting the values, we get,

$2.5 \times 101325 = 1.67 \times 10^{8} \times χ_{C O_{2}}$

$\Rightarrow χ_{C O_{2}} = \frac{2.5 \times 101325}{1.67 \times 10^{8}} = 1.517 \times 10^{- 3}$

Moles of $C O_{2}$
Molar mass of water $H_{2} O$
$= 1 \times 2 + 16 \times 1$
$= 18 g m o l^{- 1}$
Mass of $500 m L$ of water
$= 500 g$ (density of water) $= 1 g c m^{- 3}$
Therefore, number of moles of water $(H_{2} O)$ in $500 m L$ of water,
$= \frac{500}{18} = 27.78 m o l$

Now, since $n_{C O_{2}}$ is very small, we can write

$χ_{C O_{2}} = \frac{n_{C O_{2}}}{n_{H_{2} O} + n_{C O_{2}}} \approx \frac{n_{C O_{2}}}{n_{H_{2} O}}$

By putting the values, we get,

$n_{C O_{2}} = 1.517 \times 10^{- 3} \times 27.78$
$n_{C O_{2}} = 42.14 \times 10^{- 3} m o l$

Mass of $C O_{2}$
Molar mass of $C O_{2}$
$= 12 + 2 \times 16$
$44 g m o l^{- 1}$
Therefore, mass of $C O_{2}$ present
$= m o l a r m a s s o f C O_{2} \times n_{C O_{2}}$
$= 44 \times 42.14 \times 10^{- 3}$
$= 1.854 g$

Final Answer:
The quantity of $C O_{2}$ in $500 m L$ of soda water is $1.854 g$ .

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Density

Standard VIII Physics

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