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Standard XII

Chemistry

Density

In a cubic cr...

Question

In a cubic crystal of CsCl (density $= 3.97$ g cm $^{- 3}$ ), the eight corners are occupied by $C l^{-}$ ions with $C s^{+}$ ions at the body centre. Calculate the distance (in $˚ A$ ) between the neighbouring $C s^{+}$ and $C l^{-}$ ions.

1

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3.57

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7.14

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None of the above

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Solution

The correct option is B $3.57$
$z = 1$ because per unit cell only one $C s C l$ is present
$d = \frac{z M}{N_{A} \times V}$
$3.97 = \frac{1 \times 168.36}{6.023 \times 10^{23} \times (a)^{3}}$
$a^{3} = 7.04 \times 10^{- 23}$
$a = 4.13 \times 10^{- 8}$
Now,

$\sqrt{3} a = 2 (r_{+} + r_{-})$
$(r_{+} + r_{-}) = 3.57 ˚ A$

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In a cubic crystal of CsCl (density =3.97 g cm−3), the eight corners are occupied by Cl− ions with Cs+ions at the body centre. Calculate the distance (in ˚A) between the neighbouring Cs+and Cl− ions.

The correct option is B 3.57z=1 because per unit cell only one CsCl is present d=zMNA×V3.97=1×168.366.023×1023×(a)3a3=7.04×10−23a=4.13×10−8Now,√3a=2(r++r−)(r++r−)=3.57˚A

In a cubic crystal of CsCl (density $= 3.97$ g cm $^{- 3}$ ), the eight corners are occupied by $C l^{-}$ ions with $C s^{+}$ ions at the body centre. Calculate the distance (in $˚ A$ ) between the neighbouring $C s^{+}$ and $C l^{-}$ ions.