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A metal has a 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 of the metal is :
( $N_{A}$ Avogadro's constant $= 6 .02 \times 10^{23} {mol}^{- 1}$ )

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$40 g {mol}^{- 1}$

$30 g {mol}^{- 1}$

$27 g {mol}^{- 1}$

${20 g mol}^{- 1}$

answer is C.

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Detailed Solution

$d = \frac{ZM}{N_{A} a^{3}} (Z = 4 for fcc) M = \frac{d \times N_{A} \times a^{3}}{Z} = \frac{2 .72 \times 6 .023 \times 10^{23} \times {(404 \times 10^{- 10})}^{3}}{4} M = 26 .99 \approx 27 g {mol}^{- 1}$

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