Determine the number of moles in O2 in a volume of 14.48 mL at 292 K and 606 torr.?

Question

Also i couldn't get this one to work out either.
What mass of air is contained in 5.36x10^5 L volume at .98 atm and 3.0x10^1 degree Celcius? The molar mass of air is 29.0 g/mol.
thanks

Answer 1

Use PV = nRT, where n is number of moles.

If R is in L atm K-1 mol-1, remember to express P in atm, and ofcourse to use degrees Kelvin.

For second problem, find n and multiply by 29.0 g/mol

Nothing more needed!

Answer 2

Use the ideal gas law PV = nRT.

PV = nRT
But first you need to convert pressure into units of atm
Pressure(in atm) = 606/760 = 0.797 atm

PV = nRT
(0.797 atm) ( 0.01448 L) = n ( 0.08260 L.atm/(K/mol)) (292 K)

n = (0.797 atm)(0.01448 L)
__________________________
( 0.08260) ( 292 K)

n = 4.8 * 10^4 mol O2

As for your second question, use the same formula as above after converting the degree Celsisus into Kelvin. Adding 273 to the given temperature in degrees Celsius will give you your temp. in Kelvin.

Now after you find the number of moles using the formula, use the following formula to find mass:

Mass = Number of moles(n) . Molar mass

Good luck with the problem!


n =