KVPY SX SB Exam 20141 Question Paper

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Question : 106 of 120

Marks: +1, -0

The radius of K

^{+}

is 133 pm and that of Cl

^{-}

is 181 pm. The volume of the unit cell of KCl expressed in 10

^{-22}

cm

^{3}

is

0.31
1.21
2.48
6.28

Solution:

r_{K ^{+}}+r_{Cl^{-}} =a/2

133 + 181 =a/2

a = 2 (133 + 181)

a = 628

pm
or

a = 628 × 10^{–10}

cm
and
volume

= a^3 = (6.28 × 10^{–8})^3 \cm^{3}

= 2.4767 × 10^{–22} \cm^3

≈ 2.48 × 10^{–22} \cm^3

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