Answers:
1) 732.1 mmHg; 2) 0.001 014 mol; 3) 0.001 014 mol; 4) 28.7 g/mol; 5) 187 ppt.
Explanation:
1) Partial pressure of hydrogen
You are collecting the gas over water, so
[tex]p_{\text{atm}} = p_{\text{H}_{2}} + p_{\text{H}_{2}\text{O}}[/tex]
[tex]p_{\text{H}_{2}} = p_{\text{atm}} - p_{\text{H}_{2}\text{O}}[/tex]
[tex]p_{\text{atm}} = \text{754.6 mmHg}[/tex]
At 24.1 °C, [tex]p_{\text{H}_{2}\text{O}} = \text{22.5 mmHg}[/tex]
[tex]p_{\text{H}_{2}} = \text{754.6 mmHg} - \text{22.5 mmHg} = \textbf{732.1 mmHg}[/tex]
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2) Moles of H₂
We can use the Ideal Gas Law.
pV = nRT Divide both sides by RT and switch
n = (pV)/(RT)
p = 732.1 mmHg Convert to atmospheres
p = 732.1/760 Do the division
p = 0.9633 atm
V = 25.67 mL Convert to litres
V = 0.025 67 L
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = 24.1 °C Convert to kelvins
T = (24.1 + 273.15 ) K = 297.25 K Insert the values
n = (0.9633 × 0.025 67)/(0.082 06 × 297.25) Do the multiplications
n = 0.02473/24.39 Do the division
n = 0.001 014 mol
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3) Moles of metal
The partial chemical equation is
M + … ⟶ H₂ + …
The molar ratio of M:H₂ is 1 mol M:1 mol H₂.
Moles of M = 0.001 014 × 1/1 Do the operations
Moles of M = 0.001 014 mol M
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4) Atomic mass of M
Atomic mass = mass of M/moles of M Insert the values
Atomic mass = 0.0291/0.001 014 Do the division
Atomic mass = 28.7 g/mol
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5) Relative deviation in ppt
Your metal must be in Group 2 because of the 1:1 molar ratio of M:H₂.
The metal with the closest atomic mass is Mg (24.305 g/mol).
Relative deviation in ppt = |Experimental value – Theoretical value|/Theoretical value × 1000
Relative deviation = |28.7 – 23.405|/23.405 × 1000 Do the subtraction
Relative deviation = |4.39|/23.405 × 1000 Do the operations
Relative deviation = 187 ppt
