Question

A lead atom has a mass of 3.14 x 10 to the negative 22nd g.How

Answer 1

Explanation:

The given data is as follows.

       Mass of a lead atom = $3.14 \times 10^{-22}$

       Volume = 2.00 $cm^{3}$

        Density = 11.3 $g/cm^{3}$

As it is mentioned that 1 cubic centimeter contains 11.3 grams of lead.

So, in 2 cubic centimeter there will be $2 \times 11.3 g = 22.6 g$ of lead atoms.

One lead atom has a mass of $3.14 \times 10^{-22}$. Therefore, number of atoms present in 22.6 g of lead will be as follows.

                $\frac{22.6 g}{3.14 \times 10^{-22}}$

                  = $7.197 \times 10^{22} atoms$

Thus, we can conclude that there are $7.197 \times 10^{22} atoms$ of lead are present.