Questions & Answers

Question

Answers

A. ${{10}^{19}}$

B. ${{10}^{23}}$

C. ${{10}^{25}}$

D. ${{10}^{29}}$

Answer
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To find the number of electrons present per cubic meter of copper we first need to find out the number of moles of copper present in the given volume. The formula to find the number of moles is:

\[\text{Number of moles = }\dfrac{\text{given weight}}{\text{atomic weight}}\]

The atomic weight is given to us in the question and we need to find the given weight of $Cu$in the specified volume.

We have been given the density as well as volume of the copper, so we can easily find the mass using the basic density formula.

\[\begin{align}

& Density\text{ = }\dfrac{mass}{volume} \\

& Mass=density\times volume \\

\end{align}\]

Now, we can substitute the given values and find the mass.

\[\begin{align}

& mass=(9\times {{10}^{3}}kg{{m}^{-3}})\times (1{{m}^{3}}) \\

& mass=9\times {{10}^{3}}kg \\

\end{align}\]

Now, the number of moles of copper present in the given volume will be:

\[\text{Number of moles = }\dfrac{\text{given weight}}{\text{atomic weight}}\]

Now, putting the values that we have:

\[\begin{align}

& \text{No}\text{. of moles = }\dfrac{9\times {{10}^{3}}}{63.5} \\

& \text{No}\text{. of moles = }1.417\times {{10}^{2}} \\

\end{align}\]

Now that we have the number of moles of $Cu$ present we can calculate the number of atoms of $Cu$ present. We know that $1mol$ of any material contains the Avogadro’s number of molecules. The Avogadro’s number is defined to be $6.022\times {{10}^{23}}$. So now we will multiply the number of moles of copper with the Avogadro’s number to get the number of atoms of $Cu$ present in $1{{m}^{3}}$ volume.

\[\begin{align}

& \text{No}\text{. of atoms = (1}\text{.417}\times \text{1}{{\text{0}}^{2}})\times (6.022\times {{10}^{23}}) \\

& \text{No}\text{. of atoms = 8}\text{.533}\times \text{1}{{\text{0}}^{28}} \\

\end{align}\]

Since each atom of $Cu$ is said to contribute only 1 free electron, the number of free electrons in $1{{m}^{3}}$ volume of copper will be the same as the number of atoms present in the same volume.

Now, rounding off the calculated value as per the options given. We get:

\[\text{No}\text{. of free electrons per cubic meter }\cong \text{ }{{10}^{29}}\]

Please do not be alarmed while rounding off the final answer, rounding off of 8 to 10 might be a huge change but at the order of magnitude of 28 and 29, it hardly makes a significant difference.

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