Question

How many moles of argon gas will be contained in a 0.450 L flask at STP?(show all work)0.02010.01610.02150.0237

Answer 1

We will assume that the gas behaves like an ideal gas. So we can apply the ideal gas law which is described with the following equation:

$\begin{gathered} PV=nRT \\ n=\frac{PV}{RT} \end{gathered}$

Where,

n is the number of moles of the gas

P is the pressure of the gas. At STP conditions the pressure is 1atm.

T is the temperature of the gas. At STP conditions the temperature is 273.15K

V is the volume of the gas, 0.450L

R is a contant, 0.08206atm.L/mol.K

Now, we replace the known data:

$\begin{gathered} n=\frac{1atm\times0.450L}{0.08206\frac{atm.L}{mol.K}\times273.15K} \\ n=\frac{1\times0.450}{0.08206\times273.15}mol \\ n=0.0201mol \end{gathered}$

Answer: There are contained 0.0201 mol of argon gas