Question

How many grams of Cl2 gas are contained in 3.00L of the gas at 100.0 Celsius under 750.0 Torr? End result must have correct number of sig figs.

Answer 1

Assuming that Cl2 behave ideally the Ideal gas law applied is

PV = nRT

Given data: P = 750.0 torr = 0.987 atm ........... (1 torr = 0.00132 atm)

V = 3.00 L

T = 100.0 ^oC = 100.0 + 273.15 = 373.15 K

Known data: R = 0.0821 L.atm/mole.K

By putting all the known data in Ideal gas equation,

0.987 atm * 3.00 L = n * 0.0821 L.atm/mole.K *373.15 K

n = (0.987 atm * 3.00 L) / (0.0821 L.atm/mole.K *373.15 K)

n = 0.097 mole.

Molar mass of Cl2 = 70.91 g/mol

Mass of 0.097 mole Cl2 = # of moles of Cl2 * Molar mass = 0.097 mole * 70.91 g/mol = 6.878 g.

Mass of Cl2 in container is 6.878 g

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