Question

A 0.630 gram sample of a metal, M, reacts completely with sulfuric acid according to:A volume of 291 mL of hydrogen is collected over water; the water level in the collecting vessel is the same as the outside level. Atmospheric pressure is 756.0 Torr and the temperature is 25 °C. The vapor pressure of water at various temperatures can be found in this table. Calculate the molar mass of the metal.

Answer 1

55.0 g/mol

Step-by-step explanation:

Step 1. Partial pressure of hydrogen

You are collecting the gas over water, so

`p_{\text{atm}} = p_{\text{H}_(2)} + p_{\text{H}_(2)\text{O}}`

`p_{\text{H}_(2)} = p_{\text{atm}} - p_{\text{H}_(2)\text{O}}`

`p_{\text{atm}} = \text{756.0 Torr}`

At 25 °C,
`p_{\text{H}_(2)\text{O}} = \text{23.8 Torr}`

`p_{\text{H}_(2)} = \text{756.0 Torr} - \text{23.8 Torr} = \text{732.2 Torr}`

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Step 2. Moles of H₂

We can use the Ideal Gas Law.

pV = nRT Divide both sides by RT

n = (pV)/(RT)

p = 732.2 Torr Convert to atmospheres

p = 732.2/760

p = 0.9634 atm

V = 291 mL Convert to litres

V = 0.291 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 25 °C Convert to kelvins

T = (25 + 273.15 ) K = 298.15 K

n = (0.9632 × 0.291)/(0.082 06 × 298.15)

n = 0.2804/24.47

n = 0.011 46 mol

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Step 3. Moles of metal

The partial chemical equation is

M + H₂SO₄ ⟶ H₂ + …

The molar ratio of M:H₂ is 1 mol M:1 mol H₂.

Moles of M = 0.011 46× 1/1

Moles of M = 0.011 46 mol M

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Step 4. Atomic mass of M

Atomic mass = mass of M/moles of M

Atomic mass = 0.630/0.011 46

Atomic mass = 55.0 g/mol