Question

A chemist prepares a solution of copper(II) fluoride (CuF2) by measuring out 0.032 µmol of copper(II) fluoride into a 500 mL volumetric flask and filling the flask to the mark with water.Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution.

Answer 1

Answer : The concentration of copper(II) fluoride in the solution is,
`6.4* 10^(-8)mole/L`

Explanation : Given,

Moles of copper(II) fluoride = 0.032 μ

mol =
`0.032* 10^(-6)mol=3.2* 10^(-8)mol`

conversion used :
`\mu mol=10^(-6)mol`

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 :

`\text{Molarity}=\frac{\text{Moles of copper(II) fluoride}* 1000}{\text{Volume of solution (in mL)}}`

Now put all the given values in this formula, we get:

`\text{Molarity}=(3.2* 10^(-8)mol* 1000)/(500mL)=6.4* 10^(-8)mole/L`

Therefore, the concentration of copper(II) fluoride in the solution is,
`6.4* 10^(-8)mole/L`