Question 145

Solution

We have given that

Length of edge, a = 3.61x10^{–8} cm

Atomic mass of cupper, M = 63.55 g/mol

Avogadro constant, NA = 6.022x10^{23} per mol

Number of atoms in unit cell of FCC, Z= 4

Use formula of density,

Question 146

Solution

Formula is Ni_{0.98}O_{1.00}

So the ration of Ni : O = 98:100

So if there are 100 atoms of oxygen then 98 atoms of Ni,

Let number of atoms of Ni^{+2} = x

Then number of atoms of Ni^{+3} = 98–x

Charge on Ni = charge on O

So that oxygen has charge –2

3(98–x) + 2x = 2 (100)

294 –3x +2x = 200

–x = – 94

x = 94

Percentage of Ni^{+2} = (atom of Ni^{+2}/total number of atoms of Ni)100

=100(94/98)x100

= 96%

Percentage of Ni^{+3}

=100 – Ni^{+2}

=100 – 96

= 4 %

Question 147

Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?

Solution

Radius, r = 0.144 nm,

Unit cell is fcc, edge of unit cell, a = ?

In face centered unit cell, diagonal of face = 4r = $\mathrm{a}\sqrt{2}$

[∵ face diagonal = $\mathrm{a}\sqrt{2}$]

Therefore, $\mathrm{a}=\frac{4\mathrm{r}}{\sqrt{2}}=\sqrt{2}\mathrm{r}$

$=2\times 1.4142\times 0.144\mathrm{nm}\phantom{\rule{0ex}{0ex}}=0.4073\mathrm{nm}.$

Length of side of unit cell,

a = 0.4073 nm.

Unit cell is fcc, edge of unit cell, a = ?

In face centered unit cell, diagonal of face = 4r = $\mathrm{a}\sqrt{2}$

[∵ face diagonal = $\mathrm{a}\sqrt{2}$]

Therefore, $\mathrm{a}=\frac{4\mathrm{r}}{\sqrt{2}}=\sqrt{2}\mathrm{r}$

$=2\times 1.4142\times 0.144\mathrm{nm}\phantom{\rule{0ex}{0ex}}=0.4073\mathrm{nm}.$

Length of side of unit cell,

a = 0.4073 nm.

Question 148

(a) What is the length of the side of the unit cell?

(b) How many unit cells are there in 1.00 cm

Solution

Given, radius of atom (r) = 125 pm

(a) For ccp structure, we know that,