Experimentally it was found that a metal oxide has a formula of M0.98 o . Metal M is present as M*+2 and M+3 in its oxide . Fraction of metal which exists as M3+ would be?

Kindly give explanation for the steps because I have already gone through the solution but not able to follow it..

Open in App

Solution

M0.98O
Let M2+ be x so that M3+ will be 0.98-x. Total charge on the compound must be zero so that

+2(x) + 3(0.98-x) - 2 = 0 (charge on oxygen atom = -2)
2x + 2.94 - 3x - 2 = 0
x = 0.94

Amount of M3+ present = 0.98 - 0.94 = 0.04

Fraction of M3+ present = 0.040/.98 = 0.0408

Suggest Corrections

Similar questions

Q. 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. Fraction of the metal which exists as

M^{3 +}

would be :

Q. 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.

%

of the metal which exists as

M^{3 +}

would be :

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.
Fraction of the metal which exists as $M^{3 +}$ would be:

Join BYJU'S Learning Program

Experimentally it was found that a metal oxide has a formula of M0.98 o . Metal M is present as M*+2 and M+3 in its oxide . Fraction of metal which exists as M3+ would be? Kindly give explanation for the steps because I have already gone through the solution but not able to follow it..

M0.98O Let M2+ be x so that M3+ will be 0.98-x. Total charge on the compound must be zero so that +2(x) + 3(0.98-x) - 2 = 0 (charge on oxygen atom = -2) 2x + 2.94 - 3x - 2 = 0 x = 0.94 Amount of M3+ present = 0.98 - 0.94 = 0.04 Fraction of M3+ present = 0.040/.98 = 0.0408

Experimentally it was found that a metal oxide has a formula of M0.98 o . Metal M is present as M*+2 and M+3 in its oxide . Fraction of metal which exists as M3+ would be?

Kindly give explanation for the steps because I have already gone through the solution but not able to follow it..

M0.98O
Let M2+ be x so that M3+ will be 0.98-x. Total charge on the compound must be zero so that

+2(x) + 3(0.98-x) - 2 = 0 (charge on oxygen atom = -2)
2x + 2.94 - 3x - 2 = 0
x = 0.94

Amount of M3+ present = 0.98 - 0.94 = 0.04

Fraction of M3+ present = 0.040/.98 = 0.0408