Question

Total voulme of the misture of 2g of helium and 7g of nitrogen under stp conditions is

Answer 1

Total volume of mixture of 2g of helium and 7g of nitrogen under stp conditions is 22.4 litres.

- No. of moles in 2g Helium = 2/4 = 0.5 moles

- No. of moles in 7g NItrogen = 7/14 = 0.5 moles

- Total no of moles = Moles of He + Moles of N2 = 0.5 + 0.5 = 1 mole

- At STP, volume of 1 mole of any gas = 22.4 litres

Answer 2

The total voulme occupied by a mixture of 2 g of helium and 7 g of nitrogen under S.T.P conditions is 16.8 L.

• Given,

Weight of helium = 2 g

Weight of nitrogen = 7 g

• Number of moles of an element is given by the formula :

Moles = Given Weight / Molecular Weight

• Molecular weight of helium (He) = 4 g

• Therefore, number of moles of He = 2 g / 4 g

= 1 / 2

= 0.5

• Molecular weight of nitrogen (N2) = 2 × 14 g = 28 g

• Therefore, number of moles of N2 = 7 g / 28 g

= 1 / 4

= 0.25

• Now, total number of moles of helium and nitrogen present in the mixture = 0.5 + 0.25 = 0.75

• According to Avogadro's law, one mole of any gas at standard temperature and pressure, occupies a volume of 22.4 L.

• Applying Avogadro's law, volume occupied by the mixture of helium and nitrogen gases, containing 0.75 moles = 0.75 × 22.4 L = 16.8 L

The required answer is 16.8 L.