A 0.239 g sample of a gas in a 100-mL flask exerts a pressure of 600 mmHg at 14 °C. What is the gas?

Expert Answers

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We can use the Ideal gas law here

`1mmHg = 1.316*10^-3 atm`

`PV = nRT`

`P=600xx1.316xx10^3 =0.79 atm`




`T=273+14 = 287K`

`0.79xx0.1 =(0.239xx10^-3xx0.08206xx287)/M`

`M=0.014 (Kg)/(mol)=14g/(mol)`

So the gas could be Nitrogen Or `N_2` .

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