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ChemistrySolid StateCubic CrystalsMedium2 minQB

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Passage / Comprehension

Barium titanate crystallises in the pervoksite structure which is a cubic lattice, barium ions occupying the corners of the unit cell, oxide ions at the face centre of the unit cell and titanium ion is present at the body centre. [Given : $43 = 7, 4.5 2 = 6.4, (2.2)^{3} = 10.6, (1.3)^{3} = 2.2, (4.6)^{3} = 97, 4 π = 12.6, N_{A} = 6 \times 1 0^{23}$ ] Given Atomic weight : $Ba = 137; Ti = 48, O = 16$ , Given radius : $r_{Ba^{2 +}} = 2.2 A^{o}; r_{Ti^{4 +}} = 1.3 A^{o}; r_{O^{2 -}} = 1.0 A^{o}$

What is the density of Barium Titanate ?

Options:

Answer:

Solution:

Formula and Mass ( $M$ ): The unit cell contains $1 B a^{2 +}$ , $1 T i^{4 +}$ , and $3 O^{2 -}$ ions ( $Z = 1$ ), giving a molar mass $M = 137 + 48 + (3 \times 16) = 233 g / m o l$ .

Edge Length ( $a$ ): From the body-center to face-center contact, $a = 2 \times (r_{T i^{4 +}} + r_{O^{2 -}}) = 2 \times (1.3 + 1.0) = 4.6 \overset{˚}{A} = 4.6 \times 1 0^{- 8} c m$ .

Volume ( $a^{3}$ ): Using the given value, $a^{3} = (4.6 \times 1 0^{- 8} c m)^{3} = 97.4 \times 1 0^{- 24} c m^{3}$ .

Density Calculation:

$De n s i t y = \frac{Z \times M}{N _{A} \times a ^{3}} = \frac{1 \times 233}{6 \times 1 0 ^{23} \times 97.4 \times 1 0 ^{- 24}} = \frac{233}{58.44} \approx 4.00g/c m^{3}$

(Option B).

Stream:JEESubject:ChemistryTopic:Solid StateSubtopic:Cubic Crystals

⏱ 2mℹ️ Source: QB

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