Question

calculate the number of fluoride ions m 0.234g of calcium fluoride
​

Answer 1

Answer:

1.129×10²⁴ CaF2

hope this helps you

Answer 2

Answer:

The total number of fluorine ions = 3.6×10²¹

Working:

Mass of given CaF₂→ 0.234 g

Molar mass of CaF₂ → 40 + (2 × 19) = 78 g

1 mol of CaF₂ → 6.022×10²³ molecules of CaF₂

78 g of CaF₂ →  6.022×10²³ molecules of CaF₂

1 g of CaF₂ → $\frac{6.022*10^{23} }{78}$ = 7.7×10²¹ molecules of CaF₂

0.234 g of CaF₂ → (7.7×10²¹) × 0.234 = 1.8×10²¹ molecules of CaF₂

We know that 1 molecule of CaF₂ contains 2F⁻ ions, so-

1.8×10²¹×2= 3.6×10²¹ fluorine ions

Conclusion:

0.234g of calcium fluoride has 3.6×10²¹ fluorine ions.