Question

An element with molar mass \(M\,kg\,mol^{-1}\) forms a face centred cubic unit cell with edge length 405 pm. If the density is \(2.7\times10^{3}\,kg\,m^{-3}\). What is its molar mass \(M\)? \((N_A=6.0\times10^{23}\,mol^{-1})\)

• A. \(2.59\times10^{-1}\)
• B. \(2.49\times10^{-2}\)
• C. \(2.69\times10^{-2}\)
• D. \(2.89\times10^{-1}\)

Hint:

For FCC crystals always use \(Z=4\). For BCC, \(Z=2\) and for simple cubic, \(Z=1\).

Answer 1

The Correct Option is C

Concept: For a cubic crystal, \[ \rho=\frac{ZM}{N_Aa^3} \] where \[ Z=4 \quad \text{(for FCC)} \] \[ \rho=\text{density} \] \[ a=\text{edge length} \] \[ M=\text{molar mass} \]

Step 1: Convert edge length into SI unit.
\[ a=405\,pm =405\times10^{-12}m \] \[ a^3=(405\times10^{-12})^3 =6.643\times10^{-29}m^3 \]

Step 2: Apply density formula.
\[ M=\frac{\rho N_A a^3}{Z} \] \[ M= \frac{(2.7\times10^3)(6\times10^{23})(6.643\times10^{-29})}{4} \] \[ M=2.69\times10^{-2}\;kg\,mol^{-1} \] \[ \boxed{M=2.69\times10^{-2}\;kg\,mol^{-1}} \]