How many grams of copper (II) fluoride, CuF2, are needed to make 5.1 liters of a 1.4M solution?

Answer: 724.71 grams

Explanation:

Volume of solution (v) = 5.1 liters

Concentration of solution (c) = 1.4M

Amount of CuF2 needed (n) = ?

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

make n the subject formula

n = c x v

n = 1.4M x 5.1 Liters

n = 7.14 moles

Since, 7.14 moles of CuF2 (n) is needed, use the molar mass of CuF2 to get the mass in grams.

The atomic masses of Copper = 63.5g;

and Fluorine = 19g

So, I CuF2 = 63.5g + (19g x 2)

= 63.5g + 38g

= 101.5g/mol

Then, apply the formula

Number of moles = mass in grams / molar mass

7.14 moles = m / 101.5 g/mol

m = 7.14 moles x 101.5 g/mol

m = 724.71g

Thus, 724.71 grams of copper (II) fluoride, CuF2, are needed to make 5.1 liters of a 1.4M solution