Answer to Question #193490 in General Chemistry for Susie

Question #193490

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


1
Expert's answer
2021-05-17T04:16:48-0400

CM = n / V

n = m / M

M (BeCl2) = 79.92 g/mol

n (BeCl2) = CM 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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