Question

volume occupied by 28g of nitrogen gas at stp

Answer 1

According to ideal gas law, at standard temperature and pressure 1 mole of a gas occupies a volume of 22.4 liters.

Therefore calculate the moles of 28 grams of nitrogen gas

moles= mass
           molar mass
mass= 28g
molar mass ,N2= 14 + 14= 28

moles= 28/28
         = 1 mole

Since 1 mole of an ideal gas occupies 22.4 liters at stp, then the volume of 1 mole of N2= 22.4 liters

Therefore volume occupied by 28g of nitrogen = 22.4 liters

Answer 2

According to ideal gas law, at standard temperature and pressure 1 mole of a gas occupies a volume of 22.4 liters.

Therefore calculate the atoms of 28 grams of nitrogen gas

Since 6.022×10^23 atoms of an ideal gas occupies 22.4 liters at stp, then the volume of 1 mole of N2= 22.4 liters

Therefore volume occupied by 28g of nitrogen = 22.4 liters

Hope it helps