First, calculate the partial pressure of H2 by considering the total pressure:
[tex]P_{\text{tot}}=P_{H2}+P_{H20}[/tex]where,
Ptot = 1.1bar
PH20 = 150Torr
Convert the previous values to atm:
Ptot = 1.1 bar = 1.085 atm
PH20 = 150 Torr = 0.197 atm
Then, for PH2 you obtain:
[tex]P_{H2}=P_{\text{tot}}-P_{H20}=1.085\text{atm-}0.197\text{atm}=0.888atm[/tex]Next, use the following equation for ideal gases:
[tex]PV=\text{nRT}[/tex]where,
V: volume = 55.65mL = 0.05565 L
R: gas constant = 0.082 atm*L/mol*K
T: temperature = 60 + 273 = 333K
P: partial pressure of H2 = 0.888atm
Solve the equation for n and replace the values of the other parameters:
[tex]n=\frac{PV}{RT}=\frac{(0.888atm)(0.05565L)}{(0.082atm\cdot\frac{L}{\text{mol}\cdot K})(333K)}=0.00181mol[/tex]Next, use the atomic weight to determine the mass of H2:
[tex]M=(\frac{2.0141g}{\text{mol}})(0.00181mol)=0.0036g[/tex]Hence, there are 0.0036 g of H2 collected above the water