Question

The ratio of masses of oxygen and nitrogen
in a particular gaseous mixture is 1 : 4. The
ratio of number of their molecule is (2014)
1) 3 : 16 2) 1:4 3) 7: 32 4) 1:8​

Answer 1

Answer :

7 : 32

Solution :

Let,

Mass of O_2 = w

Mass of N_2 = 4w

Molecules of O_2 = w/32 × N_A

Molecules of N_2 = 4w/28 × N_A

N_O2/N_N2 = w/32 × 28/4w = 7/32

Answer 2

Answer:-

Let us assume mass of O₂ in the mixture is a grams

So, mass of N₂ in the mixture = 4a grams

(as the mass ratio is 1:4)

For O₂

Molecular Mass = 32g = 1 mole

So, a gram = (1/32)×a mole = a/32 moles

number of molecules of O₂

$\implies\sf (\frac{a}{32})\times6.022\times10^{23}$

For N₂

Molecular Mass = 28g = 1 mole

A gram = (1/28)\times 4 a moles = 1/7 moles

number of molecules of N₂

$\implies\sf (\frac{a}{7})\times6.022\times10^{23}$

Therefore, the ratio of molecules of O₂ and N₂ present in the mixture

$\implies\sf \frac{[(\frac{a}{32})\times6.022\times10^{23}]}{[(\frac{a}{7})\times6.022\times10^{23}]}$

$\implies\sf (\frac{1}{32})\div (\frac{1}{7})$

$\boxed{\frac{7}{32}}$

Thus, the ratio is 7:32