Assume that each atom in a copper wire contributes one free electron. Estimate the numberof free electrons in a cpper wire having a mass of 6.4g (take the atomic weight of copper ot be 64g \({m}{o}{l}^{{-{1}}}.\)

64 grams of copper have 1 mole.

So, 6.4 grams of copper have 0.1 mole

\({N}\odot{o}{f}{a}\to{m}{s}\in{0.1}{m}{o}\le={\left({6}\times{10}^{{-{{23}}}}\times{0.1}\right)}\)

\({a}\to{m}{s}={6}\times{10}^{{22}}{a}\to{m}{s}.\)

Since 1 atoms contributes 1 electron

\({S}{o}{6}\times{10}^{{22}}{a}\to{m}{s}{c}{o}{n}{t}{r}{i}{b}{u}{t}{e}{s}{6}\times{10}^{{22}}{e}\le{c}{t}{r}{o}{n}{s}.\)

So, 6.4 grams of copper have 0.1 mole

\({N}\odot{o}{f}{a}\to{m}{s}\in{0.1}{m}{o}\le={\left({6}\times{10}^{{-{{23}}}}\times{0.1}\right)}\)

\({a}\to{m}{s}={6}\times{10}^{{22}}{a}\to{m}{s}.\)

Since 1 atoms contributes 1 electron

\({S}{o}{6}\times{10}^{{22}}{a}\to{m}{s}{c}{o}{n}{t}{r}{i}{b}{u}{t}{e}{s}{6}\times{10}^{{22}}{e}\le{c}{t}{r}{o}{n}{s}.\)

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