Question

How much positive charge is in 0.7 kg of lithium? with each atom having 3 protons and 3 electrons. The elemental charge is 1.602 × 10−19 C and Avogadro’s number is 6.023 × 1023 . Answer in units of C.

Answer 1

Explanation:

As a neutral lithium atom contains 3 protons and its elemental charge is given as $1.602 \times 10^{-19} C$. Hence, we will calculate its number of moles as follows.

          Moles = $\frac{mass}{\text{molar mass}}$

                     = $\frac{0.7 \times 1000 g}{7 g/mol}$

                     = 100 mol

According to mole concept, there are $6.023 \times 10^{23}$ atoms present in 1 mole. So, in 100 mol we will calculate the number of atoms as follows.

        No. of atoms = $100 \times 6.023 \times 10^{23}$

                               = $6.023 \times 10^{25}$ atoms

Since, it is given that charge on 1 atom is as follows.

                     $3 \times 1.602 \times 10^{-19}C$

                    = $4.806 \times 10^{-19}C$

Therefore, charge present on $6.023 \times 10^{25}$ atoms will be calculated as follows.

    $6.023 \times 10^{25} atoms \times 4.806 \times 10^{-19} C$

            $28.95 \times 10^{6}C$

Thus, we can conclude that a positive charge of $28.95 \times 10^{6}C$ is in 0.7 kg of lithium.