Answer: The micro-moles of copper (II) fluoride is [tex]30.5\mu \text{ moles}[/tex]
Explanation:
Molarity of a solution is defined as the number of moles of solute present in 1 liter of solution.
Mathematically,
[tex]Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution (in mL)}}[/tex]
We are given:
Molarity of solution = [tex]6.1\times 10^{-4}M[/tex]
Moles of solute = ? moles
Volume of the solution = 50 mL = 0.05L (Conversion factor: 1L = 1000mL)
Putting values in above equation, we get:
[tex]6.1\times 10^{-4}=\frac{\text{Moles of solute}}{0.05}\\\\\text{Moles of solute}=0.305\times 10^{-4}moles[/tex]
To convert it into micro-moles, we multiply it by [tex]10^6[/tex]
[tex]1mole=10^6\mu\text{ moles}[/tex]
Converting [tex]0.305\times 10^{-4}moles[/tex] into micro-moles:
[tex]0.305\times 10^{-4}moles=0.305\times 10^{-4}\times 10^6\\\Rightarrow 0.305\times 10^2\mu\text{ moles}=30.5\mu\text{ moles}[/tex]
Hence, the micro-moles of copper (II) fluoride is [tex]30.5\mu \text{ moles}[/tex]
