1470 cm^3 of a gas is collected over water at 303 K and 74.4 cm of Hg. If the gas weighs 1.98 g and vapour pressure of water at 30^@C is 3.2 cm of Hg, calculate the molecular weight of the gas.

1470 cm^3 of a gas is collected over water at 303 K and 74.4 cm of Hg.

1.	1470 cm^3 of a gas is collected over water at 303 K and 74.4 cm of Hg. If the gas weighs 1.98 g and vapour pressure of water at 30^@C is 3.2 cm of Hg, calculate the molecular weight of the gas.
Answer» Solution :The gas collected over water is wet. The pressure of the DRY gas can be obtained by subtracting the vapour pressure of water from the observed pressure of the wet gas. `:. ""P = 74.4 - 3.2 = 71.2 " cm Hg " (71.2)/76` atm `V = 1470 cm^3 = 1470/1000 = 1.47 L, T = 303K` If the MOLECULAR weight of the gas is M, then the number of moles in 1.98 g of gas = `(1.98)/M` According to the gas equation `PV = nRT` SUBSTITUTING the values, we have `(71.2)/76 xx 1.47 = (1.98)/M xx 0.0821 xx 303` `:. ""M = 35.8` Hence, the molecular weight of the given gas is 35.8.