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