Question

What is the ratio of volume of 1 mole of nitrogen
as to 16 g oxygen gas at STP?
1) 1 : 0.5
(2) 1:0.92
3) 0.92 : 1
(4) 0.98 : 1
​

Answer 1

Option 1, 1 : 0.5

1 mole of any Gas at S. T. P occupies 22.4 litres.

So, Volume of one mole of N2 occupies 22.4 litres.

One mole of Oxygen O2 weighs 32g,

As We know one mole occupies 22.4 litres.

We can say,

32 g of O2 occupies - - - - - - > 22.4 litres.

16 g of O2 occupies - - - - - - > x litres.

$x = 22.4 \times \frac{16}{32} \\ \\ x = \frac{22.4}{2} = 11.2$

Therefore,

Volume of one mole of N2 = 22.4 litres

Volume of 16g of Oxygen gas at STP = 11. 2 litres.

$\sf \: Ratio \: of \: Volumes \: \\ \\ = 22.4 \: \div 11.2 \\ \\ = 1 \div 0.5$

Answer 2

Answer:1:0.5

Explanation:

Ratio of Volume of Nitrogen to ratio of Volume of Oxygen will be equal to Ratio of Moles of Nitrogen to Moles of Oxygen.

As at STP, 1 mole of any ideal gas will occupy 22.4 L of Volume.

Moles of Oxygen = Mass/Molecular Weight = 16 g/32 g=0.5 mole

Therefore, Required ratio will be

= 1:0.5