Question

At a particular temperature, the solubility of Kr in water is 0.060 M when the partial pressure is 0.90 atm. What partial pressure (in atm) of Kr would give a solubility of 0.150 M

Answer 1

At a particular temperature, the solubility of He in water is 0.060 M when the partial pressure is 0.90 atm. 2.27 atm is the partial pressure of Kr would give a solubility of 0.150 M.

What is Henry's Law ?

• According to Henry's law, the weight of a gas dissolved by a liquid is proportional to the pressure of the gas onto the liquid.
• With very few exceptions, a solute molecule in an extremely diluted solution will only have solvent molecules as its close neighbors. This means that the likelihood that a specific solute molecule will escape into the gas phase is predicted to be independent of the total concentration of solute molecules.

Solution:

The solubility of gas is directly proportional to partial pressure. It is expressed as:

S = $K_ h \times P _g_a_s$

where, S = Solubility of gas

$K_h$ = Henry's Law constant

$P_g_a_s$ = Partial pressure of gas

Now, put the values in above expression we get

0.060M = $K_h$ × 0.9 atm

   $K_h$ =  = 0.066 M/atm

Now we have to find the partial pressure of He

0.150 M = 0.066 M/atm × $P_g_a_s$

$P_g_a_s$ = 2.27 atm

Learn more about the Henry's Law : brainly.com/question/23204201

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