Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
0 votes

At 293K temperature,the Henry's Law constants for $He,H_2,N_2$ and $O_2$ are 144 kbar,69 kbar,76 kbr and 34.86 kbar respectively. Which of the following show maximum solubility in water at a given pressure

$\begin{array}{1 1}(a)\;N_2\\(b)\;He\\(c)\;O_2\\(d)\;H_2\end{array}$

Can you answer this question?

1 Answer

0 votes
Answer: $O_2$
The dissolution of a gas in a liquid is governed by Henry’s law, according to which, at a given temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas.
The inverse of the Henry's law constant, multiplied by the partial pressure of the gas above the solution, is the molar solubility of the gas, $\rightarrow $ Solubility $\propto \large\frac{1}{\text{Henry's Constant}}$
Since $O_2$ has the lowest Henry's Law constant, it has the highest solubility.
answered Feb 14, 2014 by sreemathi.v
edited Jul 15, 2014 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App