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ladessa [460]
3 years ago
6

A chemist adds of a M copper(II) fluoride solution to a reaction flask. Calculate the mass in micrograms of copper(II) fluoride

the chemist has added to the flask. Round your answer to significant digits.
Chemistry
1 answer:
anygoal [31]3 years ago
7 0

The question is incomplete, here is the complete question.

A chemist adds 345.0 mL of a 0.0013 mM (MIllimolar) copper(II) fluoride CuF_2 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, 0.0013mmol\times \frac{1mol}{1000mmol}=1.3\times 10^{-6}mol

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We are given:

Moles of copper (II) fluoride solution = 1.3\times 10^{-6}mol

Molar mass of copper (II) fluoride = 101.5 g/mol

Putting values in above equation, we get:

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

Converting this into milligrams, we use the conversion factor:

1 g = 1000 mg

So,

\Rightarrow 1.32\times 10^{-4}g\times (\frac{1000mg}{1g})=0.13mg

Therefore, the mass of copper(II) fluoride is, 0.13 mg

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