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**Hint:**First find the mass of the unit cell using formula-

Mass of a unit cell = $\dfrac{{ZM}}{N}$ where Z is the effective number of atoms in the unit cell, M is molar mass; N is the Avogadro number or number of atoms. Then calculate the density by putting the given values in the following formula-

Density= $\dfrac{{{\text{Mass of unit cell}}}}{{{\text{Volume of unit cell}}}}$

**Step-by-Step Solution-**

Given, Molar Mass M= $200$g

Number of atoms N= $24 \times {10^{23}}$

Edge length a= $200$pm

Element having FCC structure has total $4$ atoms in a unit cell. The effective number of atoms in unit cell Z= $4$

So the mass of the unit cell is equal to the mass of the $4$atoms.

Then we know that Mass of a unit cell is given by= $\dfrac{{ZM}}{N}$ where Z is the effective number of atoms in unit cell, M is molar mass, N is the Avogadro number or number of atoms.

On putting the values of the formula we get,

Mass of unit cell= \[4 \times \dfrac{{200}}{{24 \times {{10}^{23}}}}\]

On solving we get,

Mass of unit cell= $3.33 \times {10^{ - 22}}$ g--- (i)

Now the volume of the unit cell= \[{\left( a \right)^3}\]

On putting the values we get,

Volume of unit cell= ${\left( {200 \times {{10}^{ - 10}}} \right)^3}$ $c{m^3}$

Then on solving we get,

Volume of unit cell= $8 \times {10^{ - 24}}$ $c{m^3}$--- (ii)

Now we have to calculate density

And we know that Density= $\dfrac{{{\text{Mass of unit cell}}}}{{{\text{Volume of unit cell}}}}$

On putting the values from eq. (i) and (ii) in the formula we get,

Density= $\dfrac{{3.33 \times {{10}^{ - 22}}}}{{8 \times {{10}^{ - 24}}}}$

On solving we get,

Density = $0.416 \times {10^2}$

Density= $41.6gc{m^{ - 3}}$

**Answer-Hence the correct answer is B.**

**Note:**In FCC structure, following points are to be noted-

- The atoms in a unit cell are all present in the corners of the crystal lattice.

- One atom is present at the centre of every face of the cube

- This face centered atom is shared between two adjacent units’ cells in the crystal lattice.

- Only half of each atom belongs to the unit cell.

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