Question

The density of solid Ag is 10.5 g/cm3. How many atoms are present per cubic centimeter of Ag?

Answer 1

Number of moles = \(\cfrac{\text{Given mass}}{\text{Molar Mass}} \)


Molar mass of Ag = 107.8 u

As density = 10.5 g/cm^3
That is , for 1 cm^3 , the mass of Ag is 10.5 grams.

So as we are given in the question to find the no. of atoms for Ag per cubic centimetres. So, we have to consider 1 cm^3 as volume.

Therefore , given mass of Ag = 10.5 gm.

So no. of moles = \(\cfrac{10.5 }{107.8}\)

Now, as we know :

Number of atoms of Ag = number of moles * Avogadro Number

[Avogadro Number = \(6.022 * 10^{23} \) ]

Solve it now.