Two elements A and B have atomic masses 40 and 60 respectively. If 40 form hcp arrangements and A is occupying 13 of the tetrahedral voids, then what will be number of hexagonal primitive unit cells w.r.t. B atoms in 130 mg of the substance? (Given :NA=6×1023

1 mole A2B3=6×1023260g of A2B3=6×10231g=6×1023260130mg=6×1023260×1301000=3×1020with respect to B=3×1020×32 =4.5×1020

Two elements A and B have atomic masses 40 and 60 respectively. If 40 form hcp arrangements and A is occupying 13 of the tetrahedral voids, then what will be number of hexagonal primitive unit cells w.r.t. B atoms in 130 mg of the substance? (Given :NA=6×1023

1 mole A2B3=6×1023260g of A2B3=6×10231g=6×1023260130mg=6×1023260×1301000=3×1020with respect to B=3×1020×32 =4.5×1020

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Two elements A and B have atomic masses 40 and 60 respectively. If 40 form hcp arrangements and A is occupying $\frac{1}{3}$ of the tetrahedral voids, then what will be number of hexagonal primitive unit cells w.r.t. B atoms in 130 mg of the substance? (Given : $N_{A} = 6 \times 10^{23})$

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$1.0 \times 10^{20}$

$4.5 \times 10^{20}$

$4 \times 10^{20}$

$1.2 \times 10^{21}$

answer is B.

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$\begin{array}{l} 1 mole A_{2} B_{3} = 6 \times 10^{23} \\ 260 g of A_{2} B_{3} = 6 \times 10^{23} \\ 1 g = \frac{6 \times 10^{23}}{260} \\ 130 mg = \frac{6 \times 10^{23}}{260} \times \frac{130}{1000} = 3 \times 10^{20} \end{array}$

with respect to $B = 3 \times 10^{20} \times \frac{3}{2}$ $= 4.5 \times 10^{20}$

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Two elements A and B have atomic masses 40 and 60 respectively. If 40 form hcp arrangements and A is occupying 13 of the tetrahedral voids, then what will be number of hexagonal primitive unit cells w.r.t. B atoms in 130 mg of the substance? (Given :NA=6×1023

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Two elements A and B have atomic masses 40 and 60 respectively. If 40 form hcp arrangements and A is occupying $\frac{1}{3}$ of the tetrahedral voids, then what will be number of hexagonal primitive unit cells w.r.t. B atoms in 130 mg of the substance? (Given : $N_{A} = 6 \times 10^{23})$