The number of formula units of calcium fluoride CaF₂ present in 146.4 g of CaF₂(The molar mass of CaF₂ is 78.08 g/mol) is

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$1.129 \times 10^{24} {CaF}_{2}$

$1.146 \times 10^{24} {CaF}_{2}$

$7.808 \times 10^{24} {CaF}_{2}$

$1.877 \times 10^{24} {CaF}_{2}$

answer is A.

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Detailed Solution

$Number of moles of {CaF}_{2} = \frac{146 .4}{78 .08} = 1 .875$
$∴ No. of formula units = 1 .875 \times N_{A}$
$= 1.875 \times 6.023 \times 10^{23} = 11.29 \times 10^{23} = 1.129 \times 10^{24}$

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