A 23.6g sample of an unknown gas occupies a volume of 12.0 L at standard temperature and pressure. What is the molecular mass of this gas?

STP conditions are 1 atm pressure and 273 K temperature

PV = nRT

P = 1 atm

V = 12 L

n = ? moles

R = 0.0821 L atm/K/mole

T = 273 K

1 X 12 = n X 0.0821 X 273

12 = n X 22.413

n = 12/22.413 = 0.535

no.of moles = amount of gas/molar mass of gas = 0.535

23.6 / molar mass of gas = 0.535

molar mass of gas = 23.6/0.535 = 44.11 g/mole

thats the molar mass of CO2

STP conditions are 1 atm pressure and 273 K temperature

PV = nRT

P = 1 atm

V = 12 L

n = ? moles

R = 0.0821 L atm/K/mole

T = 273 K

1 X 12 = n X 0.0821 X 273

12 = n X 22.413

n = 12/22.413 = 0.535

no.of moles = amount of gas/molar mass of gas = 0.535

23.6 / molar mass of gas = 0.535

molar mass of gas = 23.6/0.535 = 44.11 g/mole

thats the molar mass of CO2

Explanation:

nuclear fission produces elements that are heavier than helium.

the elements that are used in nuclear fission and their products are much heavier than helium.