Answer: The amount of hydrogen gas collected will be 0.073 g
Explanation:
We are given:
Vapor pressure of water = 23.78 mmHg
Total vapor pressure = 740 mmHg
Vapor pressure of hydrogen gas = Total vapor pressure - Vapor pressure of water = (740 - 23.78) mmHg = 716.22 mmHg
To calculate the amount of hydrogen gas collected, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 716.22 mmHg
V = Volume of the gas = 0.947 L
T = Temperature of the gas = [tex]25^oC=[25+273]K=298K[/tex]
R = Gas constant = [tex]62.364\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
n = number of moles of hydrogen gas = ?
Putting values in above equation, we get:
[tex]716.22mmHg\times 0.947L=n\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{716.22\times 0.947}{62.364\times 298}=0.0365mol[/tex]
To calculate the mass from given number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of hydrogen gas = 0.0365 moles
Molar mass of hydrogen gas = 2 g/mol
Putting values in above equation, we get:
[tex]0.0365mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.0365mol\times 2g/mol)=0.073g[/tex]
Hence, the amount of hydrogen gas collected will be 0.073 g