One weighs a glass jar 3 times: First time: Without any air at all Second time:

Question

One weighs a glass jar 3 times:

First time: Without any air at all
Second time: Filled with air at normal pressure (1 atm)
Third time: Filled with an unknown gas at the pressure 1.5 atm
The masses are equal to 200, 204 and 210g. The temperature is the same in all cases.

Decide:
a) The mol mass of the unknown gas
b) Which gas(es) from the table it may be
Assume that the air is an ideal gas, its molecular weight is 29 g / mole and the atmospheric pressure is 100 kPa.

C2H2 l etyn eller acetvlen 2 luft 3 koldioxid 4 kolmonoxid 5 etan 6 krypton 7 syre CO CO C H6 8 ozon 9 svavelmonoxid SO 10 xeno

Explanation / Answer

weight of glass jar alone = 200g

weight of jar with air = 204g

weight of air = 4g

PV = nRT

=> (1atm)(V) = (4/29)RT ..............(1)

weight of jar with unknown gas = 210g

weight of unknown gas = 10g

PV = nRT

=> (1.5atm)(V) = (10/M)RT .................(2)

where, M is the molecular weight of the unknown gas

(1) divided by (2) we get,

1/1.5 = 4M/290

=> M = 48.34g

Molecular weight of ozon(O3) = 48g

Therefore, the unknown gas is probable ozon

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