Question

A cylinder of compressed gas rolls off a boat and falls to the bottom of a lake. Eventually it rusts and the gas bubbles to the surface. A chemist collects a sample of the gas with the idea of trying to identify the gas. The wet gas collected occupies a volume of 289 mL at a pressure of 743 torr and temperature of 28.0°C. The vapor pressure of water at 28.0°C is 0.0372 atm. 1. Calculate the volume (L) that the gas occupies after it is dried (the water vapor removed) and stored at STP. Submit Answer Tries 0/99 The mass of the dry gas is 222 mg. A fragment of the tank indicates that the gas is a monoatomic element. 2. Write out the full name of the gas. Submit Answer Tries 0/99

Answer 1

The daltons law states that the vapour pressure of the wet gas is sum of vapour pressure of dry gas and vapour pressure of water at the same temperature .

Aus:-1) Given Volume of gas collected = 289 mL ressure of gas = 743 torr Temperature of gas = 28°c. we know that 1 atm = 760 tor. & 743 torr = 0.977 atro Hence P = 0.977 atm. now from Dalton's law of partial pressure the total pressure of gas is given as :- b = Parry t Pro now Pressure of dry gas is as :- Pary = P = P = 0.977 -0.0372 I low = 0.940 atm J --- now the pressure of the gas changed. hence its volume also reduced now we also know that at constant temperature : - PV = v. comparing the pressure & volume of

0.977 x 289 = 0.940 x v. V2 = 0.977 x 289 = 300.37 ml 0.940 - Hence volume of dry gas = 300.37 mL 2) now moles of dry ges from ideal gas low PV = ORT De Rv - 0.940 x 300:37 RT 0.082 x 301 x loo Here R = 0.082 L alm mol-k Hence n = 11.43 x 10 we know that no. of moles = man molar mass Hence Given mass of dry gas = 222 mg Hence molar man = 222 x 1 g - 11.43 x 103 mol . Molar mass = 19.40 g/mol Since gas is monoatomic hence it is Hence it is Neon), as its molar mass is nearly 20 -

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