The K_H values (in ktext{bar}) of argon (I),c | vedclass

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

Question

(C) According to Henry's law,the solubility of a gas in a liquid is inversely proportional to its Henry's law constant $(K_H)$.Mathematically,$	ext{S

Vedclass · Accepted Answer

$I < II < IV < III$

Vedclass · Answer

(C) According to Henry's law,the solubility of a gas in a liquid is inversely proportional to its Henry's law constant $(K_H)$.Mathematically,$	ext{Solubility} \propto \frac{1}{K_H}$.Given $K_H$ values at $298 \ K$:Argon $(I)$: $40.3 \ k	ext{bar}$Carbon dioxide $(II)$: $1.67 \ k	ext{bar}$Methane $(IV)$: $0.413 \ k	ext{bar}$Formaldehyde $(III)$: $1.83 	imes 10^{-5} \ k	ext{bar}$Comparing the $K_H$ values: $1.83 	imes 10^{-5} Therefore,the order of $K_H$ is $III Since solubility is inversely proportional to $K_H$,the increasing order of solubility is $I < II < IV < III$.

The $K_H$ values (in $k\text{bar}$) of argon $(I)$,carbon dioxide $(II)$,formaldehyde $(III)$,and methane $(IV)$ are $40.3$,$1.67$,$1.83 \times 10^{-5}$,and $0.413$ respectively at $298 \ K$. The increasing order of solubility of these gases in liquid is:

Similar Questions

Dry air contains $79 \% \,N_2$ and $21 \% \,O_2$. At $T \,K$, if Henry's law constants for $N_2$ and $O_2$ in water are $8.57 \times 10^4 \,atm$ and $4.56 \times 10^4 \,atm$ respectively, the ratio of mole fractions of $N_2$ and $O_2$ dissolved in water at $1 \,atm$ is

Calculate the pressure of the gas if the solubility of the gas in water at $25^{\circ} C$ is $6.85 \times 10^{-4} \ mol \ dm^{-3}$. (Henry's law constant $K_H$ is $6.85 \times 10^{-4} \ mol \ dm^{-3} \ bar^{-1}$) (in $bar$)

Solubility of which among the following solids in water changes slightly with temperature?

The solubility of $CO_2$ in water increases with

State Henry's law and explain it. Also,mention its applications.

Vedclass Test Series

Exam Paper Generator

Online Exam Module