Question

An alloy has Fe, Co, and Mo equal to 71%, 12% and17% respectively. How many cobalt atoms are there in a cylinder of radius 2.50cm and a length of 10.0cm?
[The density of alloy is $8.20g \cdot mL$. The atomic weight of cobalt =58.9]
(A) $0.198 \times {10^{23}}$
(B) $198 \times {10^{23}}$
(C) $19.8 \times {10^{23}}$
(D) $1.98 \times {10^{23}}$

Answer 1

Hint: The formula to find the volume of the cylinder is
\[V = \pi {r^2}h\]
The formula of density is
\[{\text{Density = }}\dfrac{{{\text{Weight}}}}{{{\text{Volume}}}}\]

Complete step by step solution:
We need to find the weight of the cobalt present in the alloy in order to find the number of its atoms present. We will find the weight of cobalt in the alloy by using its density.
- We are given that the cylinder is having a radius of 2.50cm and length of 10.0cm. We know that the volume of the cylinder is
\[V = \pi {r^2}h\]
We know that r is the radius and h is the length of the cylinder. So, putting the available values in the above equation will give
\[V = (3.14){(2.50)^2}(10) = (3.14)(6.25)(10) = 196.25c{m^3}\]
But the density given has volume in the mL unit. The value we found of volume is in the $c{m^3}$ unit. So, we will need to convert it to mL unit.
We know that 1 $c{m^3}$ =1 mL.
So, 196.25$c{m^3}$ =196.25 mL.
Now, we will find the weight of the cylinder using its density.
\[{\text{Density = }}\dfrac{{{\text{Weight}}}}{{{\text{Volume}}}}\]
Putting all the available values in the above equation, we get
\[8.20 = \dfrac{{{\text{Weight}}}}{{196.25}}\]
So, we can write that
\[{\text{Weight = 8}}{\text{.20}} \times {\text{196}}{\text{.25 = 1609}}{\text{.25g}}\]
So, we obtained that weight of the cylinder. Now, we are given that cobalt makes 12% of the alloy. So, the weight of cobalt in the alloy will be $\dfrac{{12 \times 1609.25}}{{100}} = 193.11g$
Now, we will find the number of atoms of cobalt present in 193.11g of cobalt.
We know that 58.9g (Atomic weight) of cobalt has $6.023 \times {10^{23}}$ atoms. So, 193.11gm of cobalt will have $\dfrac{{193.11 \times 6.023 \times {{10}^{23}}}}{{58.9}} = 19.75 \times {10^{23}}$ atoms.
Thus, we can conclude that there will be $19.75 \times {10^{23}}$ atoms will be present in the alloy.

So, the correct answer is (C).

Note: Remember that one mole of any compound or species includes $6.023 \times {10^{23}}$ number of species. This number was given by Avogadro and so it is also known as Avogadro number. The atomic weight of any atom is the weight of one mole of atoms