Silver (atomic weight `108 g mol^(-1))` has a density of `10.5 g cm^(-3)`. The number of silver atoms on a surfaces of area `10^(-12) m^(2)` can be expressed in scientific notation as `Y xx 10^(-x)`, The value of `x` is …….

Answer

7

Explanation

`d=("mass")/(V) rArr 10.5 g//cc" means in 1 cc"rArr 10.5g` of Ag is present
Number of atoms of Ag in 1 cc `rArr (10.5)/(108) xx N_(A)`
In 1 cm, number of atoms of Ag `=sqrt((10.5)/(108) N_(A))`
In `1 cm^(2)`, number of atoms of Ag `=((10.5)/(108) N_(A))^(2//3)`
In `10^(-12)m^(2) or 10^(-8) cm^(2),` number of atoms of
`Ag=((10.5)/(108) N_(A))^(2//3) xx 10^(-8) =((1.05 xx 6.022 xx 10^(24))/(108))^(2//3) xx 10^(-8)`
Hence x=7 `=1.5 xx 10^(7)`