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, $6.4\times 10^{-8}mole/L$

Explanation : Given,

Moles of copper(II) fluoride = 0.032 μ

mol = $0.032\times 10^{-6}mol=3.2\times 10^{-8}mol$

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 :

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

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

$\text{Molarity}=\frac{3.2\times 10^{-8}mol\times 1000}{500mL}=6.4\times 10^{-8}mole/L$

Therefore, the concentration of copper(II) fluoride in the solution is, $6.4\times 10^{-8}mole/L$