Question

a sample of hydrogen (h2) gas was collected over water at 60°c. if the total volume of gas collected was 50.4 ml and the atmospheric pressure was 1.08 bar, what mass (in g) of hydrogen gas was collected above the water? the vapour pressure of water at 60°c is 150 torr.

Explanation:

Step1: Convert pressure units

1 bar = 750.062 torr. So, 1.08 bar = 1.08×750.062 torr = 810.067 torr. The partial - pressure of hydrogen, $P_{H_2}$, is the atmospheric pressure minus the vapor - pressure of water. $P_{H_2}=810.067 - 150$ torr = 660.067 torr. Convert to atm: $P_{H_2}=\frac{660.067}{760}$ atm $\approx0.8685$ atm.

Step2: Convert volume to SI unit

The volume $V = 50.4$ mL = 0.0504 L.

Step3: Use ideal gas law ($PV = nRT$)

The temperature $T=(60 + 273.15)$ K = 333.15 K, and $R = 0.0821$ L·atm/(mol·K). From $PV=nRT$, we can solve for the number of moles $n$ of hydrogen: $n=\frac{P_{H_2}V}{RT}=\frac{0.8685\times0.0504}{0.0821\times333.15}$ mol.
$n=\frac{0.0438724}{27.351615}$ mol $\approx0.001604$ mol.

Step4: Calculate the mass of hydrogen

The molar mass of hydrogen ($H_2$) is $M = 2.016$ g/mol. The mass $m$ of hydrogen is $m=nM$. So, $m = 0.001604\times2.016$ g $\approx0.00323$ g.

Answer:

0.00323 g