Question

A chemist adds of a 0.0013 mM 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.

Answer 1

The mass of copper(II) fluoride is 45.54 micrograms

Step-by-step explanation:

A chemist adds 345.0 mL of a 0.0013 mM 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.

Step 1: Data given

Molarity of the copper(II) fluoride solution = 0.0013 mM =

Volume of the solution 345.0 mL = 0.345 L

Molar mass copper(II) fluoride = 101.54 g/mol

Step 2: Calculate moles of copper(II) fluoride

Moles CuF2 = molarity * volume

Moles CuF2 = 0.0000013 M * 0.345 L

Moles CuF2 = 0.0000004485 moles

Step 3: Calculate mass of CuF2

Mass CuF2 = moles * Molar mass

Mass CuF2 = 0.0000004485 moles * 101.54 g/mol

Mass CuF2 = 0.00004554 grams = 0.04554 miligrams = 45.54 micrograms

The mass of copper(II) fluoride is 45.54 micrograms

Answer 2

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

Step-by-step explanation:

We are 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* (1mol)/(1000mmol)=1.3* 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* 10^(-6)mol`

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

Putting values in above equation, we get:

`1.3* 10^(-6)mol=\frac{\text{Mass of copper (II) fluoride}}{101.5g/mol}\\\\\text{Mass of copper (II) fluoride}=(1.3* 10^(-6)mol* 101.5g/mol)=1.32* 10^(-4)g`

Converting this into milligrams, we use the conversion factor:

1 g = 1000 mg

So,

`\Rightarrow 1.32* 10^(-4)g* ((1000mg)/(1g))=0.13mg`

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