Calculate molar mass of 1.29L gas which at 18 C and 765 torr weigh 2.7 grams?

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According to the combined gas law:

PV = n RT = g/M RT

Where, P is the pressure of the gas, V its volume, g is the mass of the gas in grams, M its molar mass, R is the universal gas constant (=0.08205 lit-atm/K) and T is the absolute temperature.

Let, the molar mass of the gas be M. The pressure 765 torr = 765/760 atmospheres. The temperature 18°C = (273+18) = 291K. Putting th` ` e values we get,

`765/760*1.29 = 2.7/M*0.08205*291`

Or, `M =(2.7*0.08205*291*760)/(765*1.29)`

`= 49.65`

Therefore, molar mass of the gas is 49.65.

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