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.

Respuesta :

Explanation:

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

          Moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                     = [tex]\frac{0.7 \times 1000 g}{7 g/mol}[/tex]

                     = 100 mol

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

        No. of atoms = [tex]100 \times 6.023 \times 10^{23}[/tex]

                               = [tex]6.023 \times 10^{25}[/tex] atoms

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

                     [tex]3 \times 1.602 \times 10^{-19}C[/tex]

                    = [tex]4.806 \times 10^{-19}C[/tex]

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

    [tex]6.023 \times 10^{25} atoms \times 4.806 \times 10^{-19} C[/tex]

            [tex]28.95 \times 10^{6}C[/tex]

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