Q. The number of formula units of calcium fluoride, ${CaF}_{2}$ present in 146.4 g of ${CaF}_{2}$ (the molar mass of ${CaF}_{2}\$is 78.08 g/mol) is
A) $1.129 \times 10^{24}\ {CaF}_{2}$
B) $1.146 \times 10^{24}\ {CaF}_{2}$
C) $7.808 \times 10^{24}\ {CaF}_{2}\$
D) $1.877 \times 10^{24}\ {CaF}_{2}$
Explanation
(a)
${CaF}_{2} = 146.4\ g$
Molecular weight of ${CaF}_{2} = 78.08g/mol$
Moles of ${CaF}_{2} = \frac{\ weight}{molecular\ weight}$
=$\ \frac{146.4}{78.08} = 1.875\ mol$
Number of formula units of
${CaF}_{2}\ in$ 146.4 g of ${CaF}_{2}$
$= No.of\ moles \times 6.022 \times 10^{23}$
$= 1.875 \times 6.022 \times 10^{23}$
$= 11.29 \times 10^{23}$
$= 1.129 \times 10^{24}\ {CaF}_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.