Question

Experimentally it was found that a metal oxide has formula $M_{0.98}O.$ Metal M, is present as $M^{2 +}$and $M^{3 +}$in its oxide. Fractions of the metal which exists as $M^{3 +}$would be

• A. $5.08\%$
• B. $7.01\%$
• C. $4.08\%$
• D. $6.05\%$

Answer 1

Answer: (C)

$4.08\%$

Answer 2

Let the fractions of metal which exists as $M^{3 +}$be x.

Then the fraction of metal as

$M^{2 +} = (0.98 - x)$

$\therefore$ $3x + 2(0.98 - x) = 2$

$x + 1.96 = 2$

$x = 0.04$

$\therefore\% ofM^{3 +} = \frac{0.04}{0.98} \times 100 = 4.08\%$