Respuesta :
The question is incomplete, here is the complete question.
A chemist adds 345.0 mL of a 0.0013 mM (MIllimolar) copper(II) fluoride [tex]CuF_2[/tex] solution to a reaction flask.
Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
Answer : The mass of copper(II) fluoride is, 0.13 mg
Explanation : Given,
Millimolarity of copper (II) fluoride = 0.0013 mM
This means that 0.0013 millimoles of copper (II) fluoride is present in 1 L of solution
Converting millimoles into moles, we use the conversion factor:
1 moles = 1000 millimoles
So, [tex]0.0013mmol\times \frac{1mol}{1000mmol}=1.3\times 10^{-6}mol[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We are given:
Moles of copper (II) fluoride solution = [tex]1.3\times 10^{-6}mol[/tex]
Molar mass of copper (II) fluoride = 101.5 g/mol
Putting values in above equation, we get:
[tex]1.3\times 10^{-6}mol=\frac{\text{Mass of copper (II) fluoride}}{101.5g/mol}\\\\\text{Mass of copper (II) fluoride}=(1.3\times 10^{-6}mol\times 101.5g/mol)=1.32\times 10^{-4}g[/tex]
Converting this into milligrams, we use the conversion factor:
1 g = 1000 mg
So,
[tex]\Rightarrow 1.32\times 10^{-4}g\times (\frac{1000mg}{1g})=0.13mg[/tex]
Therefore, the mass of copper(II) fluoride is, 0.13 mg
