Answer to Question #192777 in General Chemistry for Nixon

Question

How many grams of beryllium chloride (BeCl₂) are needed to make 250mL of a 1.5M solution?

Answer 1

C_M = n / V

n = m / M

M (BeCl₂) = 79.92 g/mol

n (BeCl₂) = C_M x V = 1.5 x 0.25 = 0.375 mol

m (BeCl₂) = n x M = 0.375 x 79.92 = 29.97 g