Calculate the quantity of CO_2 required to pr | vedclass

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Question

(C) According to Henry's law, $P = K_H 	imes x$, where $x$ is the mole fraction of $CO_2$. Given $P = 2 \,atm = 2 	imes 1.01325 	imes 10^5 \,Pa = 2

Vedclass · Accepted Answer

$2.9$

Vedclass · Answer

(C) According to Henry's law, $P = K_H 	imes x$, where $x$ is the mole fraction of $CO_2$. Given $P = 2 \,atm = 2 	imes 1.01325 	imes 10^5 \,Pa = 2.0265 	imes 10^5 \,Pa$. $K_H = 1.67 	imes 10^8 \,Pa$. $x = \frac{P}{K_H} = \frac{2.0265 	imes 10^5}{1.67 	imes 10^8} \approx 1.213 	imes 10^{-3}$. Since the amount of $CO_2$ is small, the number of moles of water in $1 \,L$ $(1000 \,g)$ is $n_{H_2O} = \frac{1000}{18} \approx 55.55 \,mol$. $x = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \approx \frac{n_{CO_2}}{n_{H_2O}}$. $n_{CO_2} = x 	imes n_{H_2O} = 1.213 	imes 10^{-3} 	imes 55.55 \approx 0.0674 \,mol$. Mass of $CO_2 = n_{CO_2} 	imes 	ext{Molar mass} = 0.0674 	imes 44 \approx 2.96 \,g$. Rounding to the nearest provided option, the correct answer is $2.9 \,g$.

Gas	$K_{H} / \text{kbar}$
$Ar$	$40.3$
$CO_{2}$	$1.67$
$HCHO$	$1.83 \times 10^{-5}$
$CH_{4}$	$0.413$

Calculate the quantity of $CO_2$ required to prepare $1 \,L$ of soda water when the soda water is packed under $2 \,atm$ of $CO_2$. [Henry's law constant for $CO_2$ is $1.67 \times 10^8 \,Pa$] (in $\,g$)

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The $K_{H}$ values for some gases at the same temperature '$T$' are given below:

Gas $K_{H} / \text{kbar}$

$Ar$ $40.3$

$CO_{2}$ $1.67$

$HCHO$ $1.83 \times 10^{-5}$

$CH_{4}$ $0.413$

where $K_{H}$ is Henry's Law constant in water. The order of their solubility in water is:

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