Mole ConceptHard

Question

The number of F ions in 4.2 g AlF3 is (Al = 27, F = 19)

Options

A.0.05
B.9.03 × 1022
C.3.01 × 1022
D.0.15

Solution

Number of F ions =$\frac{4.2}{27 + 3 \times 19} \times 6.02 \times 10^{23} \times 3 = 9.03 \times 10^{22}$

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