Question

Experimentally it was found that a metal oxide has formula M_0.98 O₁ . M is present as M²⁺ and M³⁺ in its oxide. Fraction of metal which exists as M³⁺ would be 

a. 7.01% 

b. 4.08% 

c. 6.05% 

d. 5.08%

Answer 1

b. 4.08%

Ratio of the numbers of M atoms to the number of 0 atoms is 

M: 0 = 0.98 : 1 = 98: 100 

Total +ve charge = Total -ve charge

Because, oxide is neutral

Let, number of M²⁺ ions = x 

so, number of M³⁺ ions = 98-x

On equating the charge, we get

2x + 3(98-x) = -2 (100) (Charge on O atoms = -200 

-x + 294 = 200 

x = 94

Number of M²⁺ ions = 94 

Number of M³⁺ ions = 4

Hence, percentage of M that exists as M³⁺ = 4/98 x 100

= 4.08%