The edge length of the unit cell of NaCl crystal lattice is 5.623 mathringA, density is 2.16 g cm-3 and the molar mass of NaCl is 58.5 g mol-1. The number of moles per unit cell is

Question

The edge length of the unit cell of NaCl crystal lattice is 5.623 mathringA, density is 2.16 g cm-3 and the molar mass of NaCl is 58.5 g mol-1. The number of moles per unit cell is (A) 4 (B) 3 (C) 1 (D) 2

Tardigrade Admin · Accepted Answer

Z = (ρ a3NA/M) = (2.16× (5.623× 10-8)3× 6.02 × 1023/58.5) = 3.95≈4

Q. The edge length of the unit cell of $N a Cl$ crystal lattice is $5.623 \overset{˚}{A}$ , density is 2.16 $g c m^{- 3}$ and the molar mass of $N a Cl$ is $58.5 g$ $m o l^{- 1}$ . The number of moles per unit cell is

4

3

1

2

$Z = \frac{ρ a ^{3} N _{A}}{M}$
$= \frac{2.16 \times ( 5.623 \times 1 0 ^{- 8} ) ^{3} \times 6.02 \times 1 0 ^{23}}{58.5}$
$= 3.95 \approx 4$

Q. The edge length of the unit cell of NaCl crystal lattice is 5.623A˚, density is 2.16 gcm−3 and the molar mass of NaCl is 58.5g mol−1. The number of moles per unit cell is

4

3

1

2

Z=Mρa3NA​​ =58.52.16×(5.623×10−8)3×6.02×1023​ =3.95≈4

Q. The edge length of the unit cell of $N a Cl$ crystal lattice is $5.623 \overset{˚}{A}$ , density is 2.16 $g c m^{- 3}$ and the molar mass of $N a Cl$ is $58.5 g$ $m o l^{- 1}$ . The number of moles per unit cell is

$Z = \frac{ρ a ^{3} N _{A}}{M}$
$= \frac{2.16 \times ( 5.623 \times 1 0 ^{- 8} ) ^{3} \times 6.02 \times 1 0 ^{23}}{58.5}$
$= 3.95 \approx 4$