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 = $6.02 \times 10^{23} {mol}^{- 1}$ )

27 {g mol}^{- 1}

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40 {g mol}^{- 1}

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20 {g mol}^{- 1}

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30 {g mol}^{- 1}

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Solution

The correct option is A $27 {g mol}^{- 1}$
$d = \frac{M}{N_{A} a^{3}} (Z = 4 for fcc) M = \frac{d \times N_{A} \times a^{3}}{Z} = \frac{2.72 \times 6.023 \times 10^{23} \times (404 \times 10^{- 10})^{3}}{4} = 27 {g mol}^{- 1}$

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Similar questions

Q. 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 metal has a fcc lattice. The edge length of the cell is 404 pm. The density of the metal is $2.72 g c m^{- 3}$ . The molar mass of the metal is
( $N_{A}$ Avogadro's constant = $6.02 \times 10^{23} m o l^{- 1}$ )

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 =6.02×10²³ mol^-1)

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 =6.02×10²³ mol^-1)

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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: (NA, Avogadro's constant = 6.02×1023 mol−1)

The correct option is A 27 g mol−1d=MNAa3 (Z=4 for fcc)M=d×NA×a3Z =2.72×6.023×1023×(404×10−10)34=27 g mol−1

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 = $6.02 \times 10^{23} {mol}^{- 1}$ )

The correct option is A $27 {g mol}^{- 1}$
$d = \frac{M}{N_{A} a^{3}} (Z = 4 for fcc) M = \frac{d \times N_{A} \times a^{3}}{Z} = \frac{2.72 \times 6.023 \times 10^{23} \times (404 \times 10^{- 10})^{3}}{4} = 27 {g mol}^{- 1}$