What partial pressure of He gas (in mm Hg) is required to maintain a solubility of 8.78×10-4 g/L in water at 25 °C? kH for He at 25 °C is 3.26×10-4 mol/L·atm.

5.12 × 10^-6 mmHg

Explanation:

So, we are given the following parameters from the question above and they are; solubilty = 8.78×10^-4 grams per litre (g/L), the Henry's law constant, KH= 3.26×10-4 mol/L·atm at the temperature = 25 °C. We can solve this question by using the Henry's law equation which is given below.

solubilty, Sb= Henry's law constant,KH × partial pressure,Pp.

So, slotting in the parameters given in the question And making the partial pressure,pp value the subject of the formula will give us our answer but there is a little problem, the solubilty unit given in the question is in grams per litre instead of it been in molarity unit. Therefore, we have to convert using the following way by dividing by the molar mass of Helium.

Hence 8.78×10^-4/ 4 = 0.0002195 mol per litre(mol/L) = 0.0002195 M.

Therefore, using the Henry's law formula, we have;

Partial pressure, Pp= solubilty, Sb/ Henry's law constant,KH.

Partial pressure, Pp= 0.0002195 M/ 3.26×10-4.

Partial pressure, Pp= 6.7 × 10^-9 atm.

We are to find our answer in mmHg.

Then 1 atm = 760 mmHg

6.7 × 10^-9 atm = x mmHg.

==> 5.12 × 10^-6 mmHg.

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)

im not exactly sure but i think its b!