Question

At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure (in atm) of He would give a solubility of 0.200 M

Answer 1

At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. 4.25 atm is the  partial pressure of He would give a solubility of 0.200 M.

What is Henry's Law ?

Henry's Law is a gas law states that at a constant temperature the amount of gas that dissolved in a liquid is directly proportional to the partial pressure of that gas.

What is relationship between Henry's Law constant and Solubility ?

The solubility of gas is directly proportional to partial pressure.

It is expressed as:

$S_{\text{gas}} = K_{H} P_{\text{gas}}$

where,

$S_{\text{gas}}$ = Solubility of gas

$K_{H}$ = Henry's Law constant

$P_{\text{gas}}$ = Partial pressure of gas

Now put the values in above expression we get

$S_{\text{gas}} = K_{H} P_{\text{gas}}$

0.080M = $K_{H}$ × 1.7 atm

$K_{H} = \frac{0.080\ M}{1.7\ \text{atm}}$

      = 0.047 M/atm

Now we have to find the partial pressure of He

$S_{\text{gas}} = K_{H} P_{\text{gas}}$

0.200 M = 0.047 M/atm × $P_{\text{gas}}$

$P_{\text{gas}} = \frac{0.200 M}{0.047\ \text{M/atm}}$

       = 4.25 atm

Thus from the above conclusion we can say that At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. 4.25 atm is the  partial pressure of He would give a solubility of 0.200 M.

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

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