Question

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. $1.129\times10^{24} CaF_{2}$
• B. $1.146\times10^{24} CaF_{2}$
• C. $7.808\times10^{24} CaF_{2}$
• D. $1.877\times10^{24} CaF_{2}$

Answer 1

Answer: (A)

$CaF _{2}=146.4\, g$
Molecular weight of $CaF _{2}=78.08 \,g / mol$
Moles of $CaF _{2}=\frac{\text { wt. }}{ mo \text {. wt. }}$
$=\frac{146.4}{78.08}=1.875\, mol$
Number of $CaF _{2}$ atoms in $146.4\, g$ of $CaF _{2}=$
No. of moles $\times 6.022 \times 10^{23}$
$=1.875 \times 6.022 \times 10^{23}$
$=11.29 \times 10^{23}$
$=1.129 \times 10^{24} CaF _{2}$