Question

3.15 g of an unknown gas at 35 °C and 1.10 atm is stored in a 1.85-L flask. What is the density of the gas? what is the molar mass of the gas?

Answer 1

ensity:
(1.05 g) / (1.45 L) = 0.724 g/L

Answer 2

`M=39.1g/mol`

`\rho=1.70g/L`

Step-by-step explanation:

Hello,

In this case, one uses the ideal gas equation:

`PV=nRT`

By solving for moles in terms of mass and molar mass one obtains:

`PV=(m)/(M)RT\\M=(mRT)/(PV) =(3.15g*0.082 (atm*L)/(mol*K) *(35+273)K)/(1.10atm*1.85L) \\M=39.1g/mol`

On the other hand, the density is easily computed as shown below:

`\rho=(m)/(V)=(3.15g)/(1.85L)=1.70g/L`

Best regards.