A metal of atomic mass 24 g mol^-1 has an hcp structure. The hexagonal unit cell contains 6 atoms and its volume is 24sqrt2 r^3 for the ideal c/a ratio. For r=160 pm, with sqrt2=1.414 and N_A=6.022x10^23 mol^-1, calculate the density in g cm^-3 (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3ss126

A metal of atomic mass $24 g mo l^{- 1}$ has an hcp structure. The hexagonal unit cell contains 6 atoms and its volume is $242 r^{3}$ for the ideal c/a ratio. For $r = 160 pm$ , with $2 = 1.414$ and $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , calculate the density in $g c m^{- 3}$ (two decimals).

Numerical answer type, answer in g/cm^3, tolerance 0.02.

V = 242 r^{3} = 24 \times 1.414 \times (1.60 \times 1 0^{- 8})^{3} = 1.390 \times 1 0^{- 22} c m^{3}

ρ = \frac{6 \times 24}{6.022 \times 1 0 ^{23} \times 1.390 \times 1 0 ^{- 22}} = 1.72 g c m^{- 3}