The pyknometer density of NaCl crystal is 2-165- 10 3 kg m -3 while its X-rays density is 2-178- 10 3 kg m -3- The fraction of the unoccupied sites in NaCl crystal is-

The correct option is D 5.96× 10 ^ 3 Fraction occupied = X ray density pyknometer densityX ray density Fraction unoccupied = 2.178× 10^3 ...

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Standard XII

Chemistry

Crystal Lattice and Unit Cell

The pyknomete...

Question

The pyknometer density of $N a C l$ crystal is $2.165 \times 10^{3} k g m^{- 3}$ while its X-rays density is $2.178 \times 10^{3} k g m^{- 3}$ . The fraction of the unoccupied sites in $N a C l$ crystal is:

5.968

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5.96 \times 10^{- 2}

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5.96 \times 10^{- 3}

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5.96 \times 10^{- 4}

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Solution

The correct option is D $5.96 \times 10^{- 3}$
Fraction occupied = $\frac{X- ray density - pyknometer density}{X- ray density}$

Fraction unoccupied $= \frac{2.178 \times 10^{3} - 2.165 \times 10^{3}}{2.178 \times 10^{3}}$

$= 5.96 \times 10^{- 3}$

Hence, option C is correct.

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Similar questions

Q. The pyknometric density of sodium chloride crystal is 2.165×10³ kgm^-3 while its X-rays density is 2.178×10³ kgm^-3. The fraction of unoccupied sites in sodium chloride crystal is

[CBSE PMT 2003]

Q. The pyknometric density of sodium chloride crystal is

2.165 \times 10^{3} k g m^{- 3}

while its X-ray density is

2.178 \times 10^{3} k g m^{- 3}

. The fraction of unoccupied sites in sodium chloride crystal is:

A metal (atomic mass $= 50 g$ ) has a body-centered cubic crystal structure. The density of metal is $5.96$ g cm $^{- 3}$ . Find the volume (in cm $^{3}$ ) of the unit cell.

Q. What is the distance between

{N a}^{+}

and

{C l}^{-}

ions in

N a C l

crystal if its density is

2.165 g {c m}^{- 3}

.[Atomic mass of

N a - 23 u, C l = 35.5 u

, Avogadro,s Number=

6.023 \times 10^{23}

]

Q. Calculate the value of Avagadro's number if the density of

N a C l

2.165

g cm

^{- 3}

and distance between

N a^{+}

and

C l^{-}

ions in

N a C l

crystal is

281

pm.

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The pyknometer density of NaCl crystal is 2.165×103kgm−3 while its X-rays density is 2.178×103kgm−3. The fraction of the unoccupied sites in NaCl crystal is:

The correct option is D 5.96×10−3Fraction occupied = X- ray density - pyknometer densityX- ray densityFraction unoccupied =2.178×103−2.165×1032.178×103=5.96×10−3Hence, option C is correct.

The pyknometer density of $N a C l$ crystal is $2.165 \times 10^{3} k g m^{- 3}$ while its X-rays density is $2.178 \times 10^{3} k g m^{- 3}$ . The fraction of the unoccupied sites in $N a C l$ crystal is:

The correct option is D $5.96 \times 10^{- 3}$
Fraction occupied = $\frac{X- ray density - pyknometer density}{X- ray density}$

Fraction unoccupied $= \frac{2.178 \times 10^{3} - 2.165 \times 10^{3}}{2.178 \times 10^{3}}$

$= 5.96 \times 10^{- 3}$

Hence, option C is correct.