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=(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=(1atm*0.450L)/(0.08206(atm.L)/(mol.K)*273.15K) \\ n=(1*0.450)/(0.08206*273.15)mol \\ n=0.0201mol \end{gathered}`

Answer: There are contained 0.0201 mol of argon gas