The pyknometric density of sodium chloride crystal is 2-165- 103 kg m-3 while its X-ray density is 2-178- 103 kg m-3- The fraction of unoccupied sites in sodium chloride crystal is-

The correct option is C 5.96× 10^ 3 Therefore volume unoccupied site in sodium chloride #160; =5.96 × 10^ 3 ...

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Standard X

Chemistry

Ionisation Energy

The pyknometr...

Question

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:

5.96 \times 10^{- 3}

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5.96

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5.96 \times 10^{- 2}

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5.96 \times 10^{- 1}

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Solution

The correct option is C $5.96 \times 10^{- 3}$
Therefore volume unoccupied site in sodium chloride $= 5.96 \times 10^{- 3}$

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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 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:

Sodium chloride crystal has FCC structure. Its density is 2.163 x 10³ kg m^-3. What will be the edge length of the unit cell cube?

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.

\sqrt{8938} \times (5.96)^{2} = ?

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The pyknometric density of sodium chloride crystal is 2.165×103 kg m−3 while its X-ray density is 2.178×103kg m−3. The fraction of unoccupied sites in sodium chloride crystal is:

The correct option is C 5.96×10−3Therefore volume unoccupied site in sodium chloride =5.96×10−3

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:

The correct option is C $5.96 \times 10^{- 3}$
Therefore volume unoccupied site in sodium chloride $= 5.96 \times 10^{- 3}$