Question

Metal has an fcc lattice. The edge length of the unit cell is $404$ pm. The density of the metal is $2.72$ g cm${}^{-3}$. The molar mass (in g mol${}^{-1}$) of the metal is:
[Avogadro's constant $=N_A=6.02\times {10}^{23}$ mol${}^{-1}$]

• A. $40$
• B. $30$
• C. $27$
• D. $20$

Answer 1

The correct option is C $27$

The formula for the density of the metal is as given below:
$d=\frac{Z\times M}{V\times N_A}$

Substituting values in the above expression, we get

$2.72=\frac{4\times M}{(4.04\times {10}^{-8}{)}^3\times 6.02\times {10}^{23}}$

$M=\frac{2.72\times (4.04{)}^3\times 6.02\times {10}^{-1}}{4}$ $=27$ g/mol

Hence, the molar mass of the metal is $27$ g/mol.

Hence, the correct option is $\text{C}$