Answer to Question #87673 in Chemistry for Daisyree

Question #87673

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

Expert's answer

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

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