Question

how many grams of beryllium chloride would you need to add to 125 mL of water to make a 0.050 molal solution

Answer 1

C_M=n/V n=m/M M (BeCl₂) = 80 g/mol

n (BeCl₂) = 0.050 x 0.125 = 0.006 mol

m (BeCl₂) = 0.006 x 80 = 0.5 g