EXPERIMENT 10 Gas Laws Unknown Number Trial 1 Trial 2 Mass of flask, boiling sto

Question

EXPERIMENT 10 Gas Laws Unknown Number Trial 1 Trial 2 Mass of flask, boiling stone, foil cap, 45?351, g and rubber band, g Mass of flask, boiling stone, foil cap, rubber band and vapor after cooling .557. Mass of unknown condensed liquid, g Water bath temperature at complete vaporization, ? Water bath temperature, K Barometric pressure, mm Hg Barometric pressure, atm Volume of flask, ml. 76 atm 14 1 Volume of flask, L Number of moles of vaporized unknown, mol Molar mass of unknown, g/mol Average molar mass of unknown, g/mol Calculations 4o.au-60.364 3618 141

Explanation / Answer

To obtain molar mass, first we obtain the number of moles with ideal gas law:

PV = nRT

n = PV / RT

Where:

P -> Pressure, V -> Volume, R -> Ideal Gas Constant, T -> Temperature

For trial 1:

n = 1 atm * 0.141L / 0.082 L atm/K mol * 361.75oK

n = 0.00475 moles

For trial 2:

n = 1 atm * 0.141L / 0.082 L atm/K mol * 369.85oK

n = 0.00465 moles

Now, with grams of liquid, we get molar mass:

For trial 1:

molar mass = 0.557 grams / 0.00475 moles = 117.263 g/mol

For trial 2:

molar mass = 0.648 grams / 0.00465 moles = 139.355 g/mol

Average molar mass = (117.263 + 139.355) / 2 = 128.309 g/mol

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