Experimentally it was found that a metal oxide has dormula `M_(0.98)O`. Metal M, present as `M^(2+) and M^(3+)` in its oxide. Fraction of the metal which exist as `M^(3+) and M^(3+)` in its oxide. Fraction of the metal which exists as `M^(3+)` would be:

Options

A. 0.0701
B. 0.0408
C. 0.0605
D. 0.0508

Answer

B

Explanation

(b) For one mole of the oxide Moles of M=0.98 Moles of `O_2=1` Let moles of `M^(3+)=x therefore` Moles of `M^(2+)=0.98-x` on balancing charge `(0.98-x)xx2+3x-2=0 x=0.04therefore %ofM^(3+)=(0.04)/(0.98).100.4.08%`