Question

hemistry 1215, Experiment #12; Determination of the Atomic Weight of an Unknown Metal, Post-lab Name ____________________________________ 1. A student who was measuring the atomic weight of a metal failed to thoroughly clean her metal ribbon leaving an oxide coat on the metal. Will the volume of hydrogen gas generated be lower or higher than the true value? 2. What volume will 5.00 g of helium gas occupy at 0.800 atmospheres and 25 o C? 3. Use Dalton’s law of partial pressures to calculate the pressure of hydrogen gas collected over water if the ambient pressure is 710 mm Hg and the temperature is 25o C. 4. A student didn’t equalize the water levels inside the eudiometer tube with that of his water bath. The water level inside the tube was 10 cm higher than the water level outside the tube. He ignored the difference and proceeded with his calculations. What effect will this error have on his calculated atomic mass? Explain your rationale.

Answer 1

From ideal gas law, volume of gas produced is equal to the number of moles of gas produced.

Actual reaction :M (s) + 2 HX (aq) → MX (aq) + H₂ (g)

Here, M is the metal.

If the metal containing an oxide results in the following reaction:

Experimental reaction :MO (s) + 2 HX (aq) → MX (aq) + H₂O (g)

Thus, the metal containing a layer of oxide results in formation of water molecules instead of water.

As the metal containing metal oxide, results in the more number of hydrogen gas. so the metal containing oxide produces more hydrogen gas so the volume of hydrogen gas is higher than the hydrogen gas produced from clean metal.

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From ideal gas,

PV=nRT

Pressure is 0.8 atm and temperature is 298.15 K, and R is gas constant (0.0821 atm L/mol K)

Number of moles = 5g/ 2 gmol^-1=2.5 mol

Volume of the gas =(2.5 mol *0.0821 atm L/mol K*298.15 K)/0.8 atm = 76.49 L.

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The vapor pressure of water at 298.15 K is 23.8 mmHg

From daltons law,

P_atm = P_hydrogen + P_{water vapor}

P_atm (atmospheric pressure) is 710 mmHg. P_{water vapor} (the partial pressure of water vapor) depends on the temperature of the water bath,

P_atm = P_hydrogen + P_{water vapor}

710 mmHg - 23.8 mmHg= P_hydrogen

P_hydrogen=686.2 mmHg.