Lithium borohydride $(L i B H_{4})$ , crystallizes in an orthorhombic system with $4$ molecules per unit cell. The unit cell dimensions are $a = 6.22 \circ A$ , $b = 5 \circ A$ , $c = 7 \circ A$ . If the molar mass of $L i B H_{4}$ is $21.76 {g mol}^{- 1}$ . The density of the crystal is:
$(N_{A} = 6 \times 10^{23})$

0.666 {g cm}^{- 3}

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0.585 {g cm}^{- 3}

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1.23 {g cm}^{- 3}

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None of these

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Solution

The correct option is A $0.666 {g cm}^{- 3}$

$Volume of the unit cell = 6.22 \times 5.00 \times 7.00 = 217.7 (\circ A)^{3}$
$Density = \frac{4 \times molar mass}{volume \times 6 \times 10^{23} \times (10^{- 8})^{3}}$
$= \frac{4 \times 21.76}{217.7 \times 6 \times 10^{23} \times (10^{- 8})^{3}}$
$\approx \frac{4 \times 1}{6 \times 1}$
$= 0.66 {g cm}^{- 3}$ .

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Q. Lithium borohydride

(L i B H_{4})

crystallizes in an orthorhombic system with

4

molecules per unit cell. The unit cell dimensions are

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˚ A

. If the molar mass of the compound is

21.76

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Lithium borohydride crystallizes in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are a = 6.8Å, b = 4.4Å and C = 7.2Å. If the molar mass is 21.76g. Density of crystal is (gm/cm³)

Lithium borohydride crystallizes in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are a = 6.8 Angstrom , b = 4.4 A and c = 7.2 angstrom . If the molar mass is 21. 76g , calculate density of Crystal

Lithium borohydride crystallizes in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are a = 6.8 $\circ A$ , b = 4.4 $\circ A$ and c = 7.2 $\circ A$ . If the molar mass is 21.76 g, calculate density of the crystal.

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Lithium borohydride (LiBH4), crystallizes in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are a=6.22 ∘A, b=5 ∘A, c=7 ∘A. If the molar mass of LiBH4 is 21.76 g mol−1. The density of the crystal is: (NA=6×1023)

The correct option is A 0.666 g cm−3 Volume of the unit cell =6.22×5.00×7.00=217.7 (∘A)3 Density =4×molar massvolume×6×1023×(10−8)3 =4×21.76217.7×6×1023×(10−8)3 ≈4×16×1 =0.66 g cm−3.