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JEE Main levelSingle correct optionDifficulty 2 of 3ss024

The number of tetrahedral voids present in 2 mol of a hexagonal close packed solid is (take $N_{A} = 6.022 \times 1 0^{23}$ )

A $1.2044 \times 1 0^{24}$
B $4.8176 \times 1 0^{24}$
C $2.4088 \times 1 0^{24}$
D $0.6022 \times 1 0^{24}$

Answer: C

Two mol of atoms is

2 N_{A}

atoms. A close-packed solid has 2 tetrahedral voids per atom, so voids

= 2 \times 2 N_{A} = 4 N_{A} = 2.4088 \times 1 0^{24}

.

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