Answer:
Aluminium ions present's in 65.5mL of 0.210
M All 3solution is Given below-:
Answer and Explanation: 1
The first step is to determine the moles of aluminum ions. This utilizes the volume, molar concentration, and subscripts from the chemical formula of aluminum (III) fluoride as shown.
65.5 mL×1 L1000 mL×0.210 mol AlF31 L×1 mol Al3+1 mol AlF3=0.013755 mol Al3+65.5 mL×1 L1000 mL×0.210 mol AlF31 L×1 mol Al3+1 mol AlF3=0.013755 mol Al3+
To calculate for the number of aluminum ions, we use Avogadro's number as shown.
0.013755 mol Al3+×6.022×1023 ions1 mol Al3+=8.28×1021 ions Al3+.
Itz the ans. of collage ok man.
