Question

How many grams of beryllium chloride (BeCl2) are needed to make 7.7 L of a 0.69 M solution?

Answer 1

419.5 g BeCl2

Step-by-step explanation:

M= mol solute/ L solution

mol solute= M * L solution

mol solute= (0.69 M) * (7.7 L)

mol solute=5.31 mol BeCl2

1 mol BeCl2 -> 79 g

5.31 mol BeCl2 -> x x= 419.5 g BeCl2