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

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) 1.129×1024 CaF2 (B) 1.146×1024 CaF2 (C) 7.808×1024 CaF2 (D) 1.877×1024 CaF2

Tardigrade Admin · Accepted Answer

CaF 2=146.4 g Molecular weight of CaF 2=78.08 g / mol Moles of CaF 2=( text wt. / mo text . wt. ) =(146.4/78.08)=1.875 mol Number of CaF 2 atoms in 146.4 g of CaF 2= No. of moles × 6.022 × 1023 =1.875 × 6.022 × 1023 =11.29 × 1023 =1.129 × 1024 CaF 2

Q. The number of formula units of calcium fluoride, $C a F_{2}$ present in $146.4 g$ of $C a F_{2}$ (the molar mass of $C a F_{2}$ is $78.08 g / m o l$ ) is

$1.129 \times 1 0^{24} C a F_{2}$

$1.146 \times 1 0^{24} C a F_{2}$

$7.808 \times 1 0^{24} C a F_{2}$

$1.877 \times 1 0^{24} C a F_{2}$

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

1.129×1024CaF2​

1.146×1024CaF2​

7.808×1024CaF2​

1.877×1024CaF2​

CaF2​=146.4g Molecular weight of CaF2​=78.08g/mol Moles of CaF2​=mo. wt. wt. ​ =78.08146.4​=1.875mol Number of CaF2​ atoms in 146.4g of CaF2​= No. of moles ×6.022×1023 =1.875×6.022×1023 =11.29×1023 =1.129×1024CaF2​

Q. The number of formula units of calcium fluoride, $C a F_{2}$ present in $146.4 g$ of $C a F_{2}$ (the molar mass of $C a F_{2}$ is $78.08 g / m o l$ ) is

$1.129 \times 1 0^{24} C a F_{2}$

$1.146 \times 1 0^{24} C a F_{2}$

$7.808 \times 1 0^{24} C a F_{2}$

$1.877 \times 1 0^{24} C a F_{2}$