Question

A chemist adds 50.0 mL of a 6.1 x 10^-4 M copper(II) fluoride solution to a reaction flask. Calculate the micromoles of copper(II) fluoride the chemist has added to the flask.

Answer 1

Answer: The micro-moles of copper (II) fluoride is
`30.5\mu \text{ moles}`

Step-by-step explanation:

Molarity of a solution is defined as the number of moles of solute present in 1 liter of solution.

Mathematically,

`Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution (in mL)}}`

We are given:

Molarity of solution =
`6.1* 10^(-4)M`

Moles of solute = ? moles

Volume of the solution = 50 mL = 0.05L (Conversion factor: 1L = 1000mL)

Putting values in above equation, we get:

`6.1* 10^(-4)=\frac{\text{Moles of solute}}{0.05}\\\\\text{Moles of solute}=0.305* 10^(-4)moles`

To convert it into micro-moles, we multiply it by
`10^6`

`1mole=10^6\mu\text{ moles}`

Converting
`0.305* 10^(-4)moles` into micro-moles:

`0.305* 10^(-4)moles=0.305* 10^(-4)* 10^6\\\Rightarrow 0.305* 10^2\mu\text{ moles}=30.5\mu\text{ moles}`

Hence, the micro-moles of copper (II) fluoride is
`30.5\mu \text{ moles}`