Question

vanadium (v) and copper (cu) are metals commonly used in engineering and construction applications. based on their average atomic weights, how much more dense would you expect copper to be compared to vanadium? to answer this problem you will need to assume that vanadium and copper atoms pack together in the metal structures in such a way that the same number of atoms of each would occupy exactly the same volume (this isnt actually true, but lets make this assumption for the purpose of this question). multiple choice 1 point copper will be 1.05 times more dense copper will be 1.10 times more dense copper will be 1.15 times more dense copper will be 1.25 times more dense dont test me yet, i am trying to gain an understanding before attempting to answer this question.

Explanation:

Step1: Recall density formula

Density $
ho=\frac{m}{V}$. If the number of atoms is the same and they occupy the same volume, density is proportional to the mass of the atoms.

Step2: Identify atomic weights

The atomic weight of Vanadium ($V$) is $50.942$ and of Copper ($Cu$) is $63.554$.

Step3: Calculate density ratio

Let the density of Vanadium be $
ho_V$ and of Copper be $
ho_{Cu}$. Since $V$ is the same and number of atoms is same (so mass is proportional to atomic - weight), $\frac{
ho_{Cu}}{
ho_V}=\frac{m_{Cu}}{m_V}=\frac{63.554}{50.942}\approx1.25$.

Answer:

Copper will be 1.25 times more dense