Question

calculate the volume occupied by 28u nitrogen gas

Answer 1

• Consider the condition at STP that is, the molar mass of an element given in u contains 1 molecule of the element.
• Given that 28 u nitrogen gas contains 1 molecule of nitrogen gas.
• One mole of nitrogen gas at STP contains 22.4 L
• One mole of nitrogen gas contains $6.022\times {10}^{23}$ molecules of nitrogen gas contains 22.4 L of nitrogen gas.
• Hence, the number of moles of nitrogen gas of 28 u containing 1 molecule of nitrogen gas

  $=\frac{28}{6.022\times {10}^{23}\times 28}$

  $=\frac{1}{6.022\times {10}^{23}}moles$ ($\because$ 1 mole of nitrogen contains $6.022\times {10}^{23}$ molecules)
  

• Hence, the volume occupied by 28 u nitrogen gas is:

  $=\frac{1}{6.022\times {10}^{23}}\times 22.4L$

  $=\frac{1}{6.022\times {10}^{23}}\times 22400mL$

$=3.719\times {10}^{-20}mLof{N}_{2}gas$

Hence the volume occupied by 28u nitrogen gas is $3.719\times {10}^{-20}mLof{N}_{2}gas$