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