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# The density of copper is $9\times {{10}^{3}}kg{{m}^{-3}}$and its atomic mass is $63.5u$. Each copper atom provides one free electron. Estimate the number of free electrons per cubic meter in copper:A. ${{10}^{19}}$B. ${{10}^{23}}$C. ${{10}^{25}}$D. ${{10}^{29}}$  Hint: You can use the basic density formula to start this problem. $Density=\dfrac{mass}{volume}$. Find the mass of the copper present and then divide by atomic mass to find the number of moles. Electrons can be found easily after we find the number of moles.

Complete step by step solution:
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:
$\text{No}\text{. of free electrons per cubic meter }\cong \text{ }{{10}^{29}}$

So, the correct answer is “Option D”.

Note: Each substance has a specific density. Generally, the density of water which is approximately about one gram per cubic centimetre is taken as the standard value for calculating value for calculating the density of substances. You should have a good idea about the different units of densities.
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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