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
