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

Answer:

1.77mg

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} \frac{mol H_{2}}{L} )*(0.315L)*(\frac{2g H_{2} }{1 mol H_{2}} )*(\frac{1000 mg H_{2}}{1 g H_{2}} )

[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.