IIT JEE Advanced 2010 Paper 2

We know that
Density =

Mass

Volume

Given that density = 10.5 g∕cm3, which means 10.5 g of Ag is present in 1 cm3.
Number of atoms of Ag present in 1 cm3 is

10.5

108

×NA
In 1 cm2, number of atoms of Ag is
(

10.5

108

×NA)2∕3
In 10−12m2 or 10−8cm2, the number of atoms of
Ag = (

10.5

108

×NA)2∕3×10−8
= (

10.5

108

×6.023×1023)×10−8 ~ 1.5×107
Hence, x = 7.