Question

Titanium oxide having formula has metal deficiency defect due to which it consist of two different oxidation states of the metal i.e. and If ratio of number of the metal ions i.e. is 1 : 24 then value of 19x is :

Answer 1

Answer:

Charge of Iron must balance the charge of oxygen.

Let the formula be Fe

x

O

2(x−

20

3x

)+3(

20

3x

)=2

⟹

20

2(17x)

+

20

9x

=2

⟹

20

43x

=2

⟹x=0.93

Formula of oxide: Fe

0.93

O

Answer 2

The value of 19x is 14.54.

Given:-

Formula of Titanium Oxide = Ti2xO3

Oxidation States of Titanium = Ti+³ and Ti+⁴

Ratio of Metal ions = 1:24

To Find:-

The value of 19x.

Solution:-

We can simply calculate the value of 19x by using these simple steps.

As

Formula of Titanium Oxide = Ti2xO3

Oxidation States of Titanium = Ti+³ and Ti+⁴

Ratio of Metal ions = 1:24

Here,

let the concentration of Titanium +3 ion (Ti+³) be y

then, concentration of Titanium+4 ion (Ti+⁴) = 2x - y

Also

$\frac{Ti+³}{Ti+⁴} = \frac{1}{24}$

$\frac{Ti+³}{Ti+⁴} = \frac{1}{24}$$\frac{Ti+³}{Ti+⁴} = \frac{1}{24} = \frac{y}{2x - y}$

$\frac{Ti+³}{Ti+⁴} = \frac{1}{24}$$\frac{Ti+³}{Ti+⁴} = \frac{1}{24} = \frac{y}{2x - y}$$\frac{1}{24} = \frac{4x - 3}{2x - (4x - 3)}$

$\frac{Ti+³}{Ti+⁴} = \frac{1}{24}$$\frac{Ti+³}{Ti+⁴} = \frac{1}{24} = \frac{y}{2x - y}$$\frac{1}{24} = \frac{4x - 3}{2x - (4x - 3)}$$\frac{1}{24} = \frac{4x - 3}{2x - 4x + 3}$

$\frac{Ti+³}{Ti+⁴} = \frac{1}{24}$$\frac{Ti+³}{Ti+⁴} = \frac{1}{24} = \frac{y}{2x - y}$$\frac{1}{24} = \frac{4x - 3}{2x - (4x - 3)}$$\frac{1}{24} = \frac{4x - 3}{2x - 4x + 3}$$\frac{1}{24} = \frac{4x - 3}{ - 2x + 3}$

$\frac{Ti+³}{Ti+⁴} = \frac{1}{24}$$\frac{Ti+³}{Ti+⁴} = \frac{1}{24} = \frac{y}{2x - y}$$\frac{1}{24} = \frac{4x - 3}{2x - (4x - 3)}$$\frac{1}{24} = \frac{4x - 3}{2x - 4x + 3}$$\frac{1}{24} = \frac{4x - 3}{ - 2x + 3}$$1( - 2x + 3) = 24(4x - 3)$

$\frac{Ti+³}{Ti+⁴} = \frac{1}{24}$$\frac{Ti+³}{Ti+⁴} = \frac{1}{24} = \frac{y}{2x - y}$$\frac{1}{24} = \frac{4x - 3}{2x - (4x - 3)}$$\frac{1}{24} = \frac{4x - 3}{2x - 4x + 3}$$\frac{1}{24} = \frac{4x - 3}{ - 2x + 3}$$1( - 2x + 3) = 24(4x - 3)$$- 2x + 3 = 96x - 72$

$\frac{Ti+³}{Ti+⁴} = \frac{1}{24}$$\frac{Ti+³}{Ti+⁴} = \frac{1}{24} = \frac{y}{2x - y}$$\frac{1}{24} = \frac{4x - 3}{2x - (4x - 3)}$$\frac{1}{24} = \frac{4x - 3}{2x - 4x + 3}$$\frac{1}{24} = \frac{4x - 3}{ - 2x + 3}$$1( - 2x + 3) = 24(4x - 3)$$- 2x + 3 = 96x - 72$$96x + 2x = 72 + 3$

$\frac{Ti+³}{Ti+⁴} = \frac{1}{24}$$\frac{Ti+³}{Ti+⁴} = \frac{1}{24} = \frac{y}{2x - y}$$\frac{1}{24} = \frac{4x - 3}{2x - (4x - 3)}$$\frac{1}{24} = \frac{4x - 3}{2x - 4x + 3}$$\frac{1}{24} = \frac{4x - 3}{ - 2x + 3}$$1( - 2x + 3) = 24(4x - 3)$$- 2x + 3 = 96x - 72$$96x + 2x = 72 + 3$$98x = 75$

$\frac{Ti+³}{Ti+⁴} = \frac{1}{24}$$\frac{Ti+³}{Ti+⁴} = \frac{1}{24} = \frac{y}{2x - y}$$\frac{1}{24} = \frac{4x - 3}{2x - (4x - 3)}$$\frac{1}{24} = \frac{4x - 3}{2x - 4x + 3}$$\frac{1}{24} = \frac{4x - 3}{ - 2x + 3}$$1( - 2x + 3) = 24(4x - 3)$$- 2x + 3 = 96x - 72$$96x + 2x = 72 + 3$$98x = 75$$x = \frac{75}{98}$

Now,

$19x = \frac{75}{98} \times 19$

$19x = \frac{75}{98} \times 19$$19x = \frac{1425}{98}$

$19x = \frac{75}{98} \times 19$$19x = \frac{1425}{98}$$19x = 14.54$

Hence, The value of 19x is 14.54.

#SPJ2