Answer : The concentration of copper(II) fluoride in the solution is, [tex]6.4\times 10^{-8}mole/L[/tex]
Explanation : Given,
Moles of copper(II) fluoride = 0.032 μ
mol = [tex]0.032\times 10^{-6}mol=3.2\times 10^{-8}mol[/tex]
conversion used : [tex]\mu mol=10^{-6}mol[/tex]
Volume of solution = 500 mL
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
[tex]\text{Molarity}=\frac{\text{Moles of copper(II) fluoride}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Molarity}=\frac{3.2\times 10^{-8}mol\times 1000}{500mL}=6.4\times 10^{-8}mole/L[/tex]
Therefore, the concentration of copper(II) fluoride in the solution is, [tex]6.4\times 10^{-8}mole/L[/tex]
