Question

How many fluoride ions are present in 65.5 ml of 0.210 m alf3 solution?

Answer 1

Hey there!:

Volume in liters of solution:

65.5 mL / 1000 => 0.0655 L

Number of moles of fluoride ions :

AlF3 = 3 atoms of F so :

n = M * V

n = 0.210  * 3 *  0.0655

n = 0.041265 moles of  fluoride ions

Therefore:

Number of ions  :

0.041265* ( 6.02*10²³ ) =  2.485 *10²² ions  

Hope that helps!

Answer 2

Answer: The number of fluorine ions present is $2.48\times 10^{22}$

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of aluminium fluoride = 0.210 M

Volume of solution = 65.5 mL

Putting values in above equation, we get:

$0.210M=\frac{\text{Moles of }AlF_3\times 1000}{65.5mL}\\\\\text{Moles of }AlF_3=\frac{0.210\times 65.5}{1000}=0.01375mol$

We are given a chemical compound having a chemical formula $AlF_3$

This compound is formed by 1 aluminium ions and 3 moles of fluorine ions.

According to mole concept:

1 mole of an ionic compound contains $6.022\times 10^{23}$ number of ions.

So, 0.01375 mole of aluminium fluoride will contain $3\times 0.01375\times 6.022\times 10^{23}=2.48\times 10^{22}$ number of fluorine ions

Hence, the number of fluorine ions present is $2.48\times 10^{22}$