Answer to Question #193490 in General Chemistry for Susie

General Chemistry

Question #193490

How many grams of BeCl2 are required to make 0.75 L of a 0.25 M solution?

Expert's answer

C_M = n / V

n = m / M

M (BeCl₂) = 79.92 g/mol

n (BeCl₂) = C_M x V = 1.5 x 0.75 = 1.125 mol

m (BeCl₂) = n x M = 1.125 x 79.92 = 89.91 g

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