Question

What mass of hydrogen gas, in mg, is dissolved in 3.15×102 mL of water when the partial pressure of hydrogen above the water is 3.6 atm at 25 °C? The Henry's constant for hydrogen gas in water at 25 °C is 7.8×10-4 M/atm.

Answer 1

1.77mg

Step-by-step explanation:

Hello,

Since the Henry's constant is eligible for
`H_(2)` (hydrogen gas), its concentration into the 315 mL of water is given by:

`C=P*H\\C=3.6 atm * 7.8x10^(-4) M/atm=2.808x10^(-3) M\\`

Now, with the following factors, the mass in grams is found as:

`m=(2.808x10^(-3) (mol H_(2))/(L) )*(0.315L)*((2g H_(2) )/(1 mol H_(2)) )*((1000 mg H_(2))/(1 g H_(2)) )</p><p>[tex]m=1.77 mg H_(2)`

Taking into account that the first parenthesis accounts for the concentration of
`H_(2)`, the second one for the volume of water, the third one for the molar mass of
`H_(2)` and the last one for the conversion from g of
`H_(2)` to mg of
`H_(2)`.

Best regards.