A copper plate of $20cm \times 10cm$ is to be plated on one side with silver of $1$ mm thickness. The number of silver atoms required for the plating is (density of silver = $10.8$g/c.c)

Density of silver = $10.8$g/c.c

Volume =$l \times b \times h$

Density =$\frac{ mass }{volume}$

No. of moles = given weight / molecular weight

No. of atoms = No. of moles $\times$ $6.023x{10}^{23}$

Detailed explanation: To find out the no. of atoms of silver, we first have to find out no. of

moles of silver.

Volume of silver required for plating:

Volume = $l \times b \times h$

= $20 x 10 x 0.1$

= $20c.c$

Mass of silver required for plating:

Density = mass / volume

$10.8$ g/c.c = mass / $20c.c$

Mass = $10.8$g/cc/$20c.c$

Mass = $10.8$ × $20$ g

No. of moles of silver:

No. of moles = given weight / molecular weight

No. of moles = $10.8 \times 20 g/108 g/moles$

No. of moles = $2$ moles

No. of silver atoms:

No. of atoms = $No. of moles \times 6.023 \times {10}^{23}$

No. of moles =$2 \times 6.023\times {10}^{23}$

No. of moles of silver atom = $1.2x{10}^{24}$

Therefore, Option (A) $1.2x{10}^{24}$ is correct.