The Henry's law constant (K_H) values of thre | vedclass

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

Question

(A) According to Henry's law,the solubility of a gas is inversely proportional to the Henry's law constant $(K_H)$ at a given pressure.Mathematically,

Vedclass · Accepted Answer

$B > C > A$

Vedclass · Answer

(A) According to Henry's law,the solubility of a gas is inversely proportional to the Henry's law constant $(K_H)$ at a given pressure.Mathematically,$	ext{Solubility} \propto \frac{1}{K_H}$.Given $K_H$ values are:$A = 145 	ext{ kbar}$$B = 2 	imes 10^{-5} 	ext{ kbar}$$C = 35 	ext{ kbar}$Comparing the $K_H$ values: $A (145) > C (35) > B (2 	imes 10^{-5})$.Since solubility is inversely proportional to $K_H$,the order of solubility will be the reverse of the order of $K_H$ values.Therefore,the order of solubility is $B > C > A$.

The Henry's law constant $(K_H)$ values of three gases $(A, B, C)$ in water are $145$,$2 \times 10^{-5}$,and $35$ kbar,respectively. The solubility of these gases in water follow the order:

Similar Questions

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 $0.159 \ mol \ dm^{-3} \ bar^{-1}$)?

Which of the following statement$(s)$ is/are correct regarding the solubility of solid $I_2$?

Solubility of $N_2$ gas in water at $25\,^{\circ}C$ and $0.78\, atm$ is $5.4 \times 10^{-4}\, M$. Calculate Henry's law constant for $N_2$ in water at $25\,^{\circ}C$.

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

Explain the equilibrium of gases in liquids.

Vedclass Test Series

Exam Paper Generator

Online Exam Module