Question

What is the molar mass and identity of a 5.28g sample of a gas that occupies 2.96 L at STP?

Answer 1

40 g/mol, Argon

Step-by-step explanation:

We can find the number of moles of the gas by using the equation of state for an ideal gas:

`pV=nRT`

where

`p=101,300 Pa` is the pressure of the gas at STP

`V=2.96 L = 2.96\cdot 10^(-3) m^3` is the volume of the gas

n is the number of moles

`R=8.314 J/mol K` is the gas constant

`T=0^(\circ)C=273 K` is the absolute temperature of the gas at STP

Solving for n, we find:

`n=(pV)/(RT)=((101,300)(2.96\cdot 10^(-3)))/((8.314)(273))=0.132 mol`

Now we can find the molar mass of the gas, which is given by

`M_m=(m)/(n)`

where

m = 5.28 g is the mass of the gas

n = 0.132 mol is the number of moles

Substituting,

`M_m=(5.28)/(0.132)=40 g/mol`

So, the gas in this problem is Argon, which has a molar mass of 40 g/mol.