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:

4.08 %

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6.05 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.08 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7.01 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $4.08 %$
The oxidation state of oxygen is -2.
Let the fraction of $M^{3 +}$ be x.
Then fraction of $M^{2 +}$ will be (0.98−x).
Now for the compound to be neutral,

3x+2(0.98−x) = 2
3x+1.96−2x =2
x = 2−1.96x
x = 0.04
So, the fraction of $M^{3 +}$ will be 4%.

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.

%

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:

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..

Join BYJU'S Learning Program

Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+ and M3+ in its oxide. Fraction of the metal which exists as M3+ would be:

The correct option is A 4.08%The oxidation state of oxygen is -2. Let the fraction of M3+ be x. Then fraction of M2+ will be (0.98−x). Now for the compound to be neutral, 3x+2(0.98−x) = 2 3x+1.96−2x =2 x = 2−1.96x x = 0.04 So, the fraction of M3+ will be 4%.

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:

The correct option is A $4.08 %$
The oxidation state of oxygen is -2.
Let the fraction of $M^{3 +}$ be x.
Then fraction of $M^{2 +}$ will be (0.98−x).
Now for the compound to be neutral,

3x+2(0.98−x) = 2
3x+1.96−2x =2
x = 2−1.96x
x = 0.04
So, the fraction of $M^{3 +}$ will be 4%.