A close-packed solid contains 0.25 mol of atoms. Using N_A=6.022x10^23 mol^-1, the number of tetrahedral voids present is

JEE Main levelSingle correct optionDifficulty 2 of 3ss079

A close-packed solid contains $0.25$ mol of atoms. Using $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , the number of tetrahedral voids present is

Tetrahedral voids

= 2 \times

atoms

= 2 \times 0.25 \times 6.022 \times 1 0^{23} = 3.01 \times 1 0^{23}