Question

An element with BCC structure has edge length of 500 pm. If it's density is 4 g cm$^{-3}$, find atomic mass of the element?

• A. 150 g mol$^{-1}$
• B. 100 g mol$^{-1}$
• C. 125 g mol$^{-1}$
• D. 250 g mol$^{-1}$

Hint:

To make calculations lightning fast under exam conditions, notice that $4 \times 125 = 500$. Then, $500 \times 10^{-24} \times 10^{23} = 50$. Finally, $\frac{50}{2} = 25$, and $25 \times 6 = 150$. Keeping powers of ten organized makes physical chemistry arithmetic highly manageable!

Answer 1

The Correct Option is A

Step 1: Note the given data.
BCC lattice (so $Z = 2$), edge $a = 500$ pm, density $\rho = 4$ g cm$^{-3}$; find atomic mass $M$.
Step 2: Write the density formula.
$\rho = \dfrac{Z \, M}{a^3 N_A}$, which rearranges to $M = \dfrac{\rho \, a^3 N_A}{Z}$.
Step 3: Convert the edge to cm.
$a = 500$ pm $= 5 \times 10^{-8}$ cm.
Step 4: Cube the edge.
$a^3 = (5 \times 10^{-8})^3 = 125 \times 10^{-24} = 1.25 \times 10^{-22}$ cm$^3$.
Step 5: Substitute the numbers.
$M = \dfrac{4 \times 1.25 \times 10^{-22} \times 6.022 \times 10^{23}}{2}$. The powers of ten give $10^{-22} \times 10^{23} = 10^{1}$, so the numerator $= 4 \times 1.25 \times 6.022 \times 10 = 301.1$.
Step 6: Divide and round.
$M = \dfrac{301.1}{2} = 150.55 \approx 150$ g mol$^{-1}$, which is option (1).
\[ \boxed{M \approx 150 \text{ g mol}^{-1}} \]