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

Answer:

40 g/mol, Argon

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:

$n=\frac{pV}{RT}=\frac{(101,300)(2.96\cdot 10^{-3})}{(8.314)(273)}=0.132 mol$

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

$M_m=\frac{m}{n}$

where

m = 5.28 g is the mass of the gas

n = 0.132 mol is the number of moles

Substituting,

$M_m=\frac{5.28}{0.132}=40 g/mol$

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