If CO_2 gas having a partial pressure of 1.67 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) According to Henry's law: $p = K_H 	imes \chi$Given: $p = 1.67 \ bar$,$K_H = 1.67 \ kbar = 1670 \ bar$.Calculating mole fraction $\chi$ of $CO_2$

Vedclass · Accepted Answer

$2.44$

Vedclass · Answer

(C) According to Henry's law: $p = K_H 	imes \chi$Given: $p = 1.67 \ bar$,$K_H = 1.67 \ kbar = 1670 \ bar$.Calculating mole fraction $\chi$ of $CO_2$: $\chi = \frac{p}{K_H} = \frac{1.67}{1670} = 0.001$.Since the amount of $CO_2$ is very small,$\chi = \frac{n_{CO_2}}{n_{H_2O}} \approx 0.001$.Number of moles of water in $1 \ L$ is $n_{H_2O} = \frac{1000 \ g}{18 \ g \ mol^{-1}} = 55.55 \ mol$.Therefore,$n_{CO_2} = 0.001 	imes 55.55 = 0.05555 \ mol$.Mass of $CO_2 = n_{CO_2} 	imes 	ext{Molar mass of } CO_2 = 0.05555 \ mol 	imes 44 \ g \ mol^{-1} \approx 2.44 \ g$.Thus,the amount of $CO_2$ dissolved is $2.44 \ g \ L^{-1}$.

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

If $CO_2$ gas having a partial pressure of $1.67 \ bar$ is bubbled through $1 \ L$ water at $298 \ K$,the amount of $CO_2$ dissolved in water in $g \ L^{-1}$ is approximately. (Henry's law constant of $CO_2$ is $1.67 \ kbar$ at $298 \ K$)

Similar Questions

The solubility curve of $KNO_3$ in water is shown below. The amount of $KNO_3$ that dissolves in $50 \ g$ of water at $40^{\circ} C$ is closest to $.... \ g$

Henry's law constant for the solubility of $N_2$ gas in water at $298\, K$ is $1.0 \times 10^5\, atm$. The mole fraction of $N_2$ in air is $0.6$. The moles of $N_2$ from air dissolved in $10$ moles of water at $298\, K$ and $5\, atm$ pressure is

The solution of sugar in water contains

What is the solubility of a gas in water at $25^{\circ} C$ if the partial pressure is $0.346 \ bar$ (Henry's law constant is $0.159 \ mol \ dm^{-3} \ bar^{-1}$)?

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module