How many grams of CAF2 are in 1.25 L of a 0.15 M solution of CaF2

The answer for the following problem is mentioned below.

Explanation:

Given:

volume of calcium fluoride (CaF) = 1.25 liters

molarity of the solution (M) = 0.15 M

To calculate:

mass of the calcium fluoride

We know;

To calculate the molarity of the solution;

M = \frac{n}{v}

where;

M represents molarity of the solution

n represents no of moles of the solution

V represents the volume of the calcium fluoride

We know,

n =   \frac{m}{M}      

Where,

m represents mass of the calcium chloride

M represents the molar mass of the solution

M = \frac{m}{M*V}

mass of calcium fluoride (CaF) =40 + (2×19) = 40 +  38 = 78 grams

0.15 = \frac{78}{M*1.25}

m = molarity × mass ×Volume

m = 0.15 × 78 × 1.25

m = 14.625 grams

Therefore the mass of the calcium fluoride is 14.625 grams.