How many grams of beryllium chloride (BeCl₂) are needed to make 250mL of a 1.5M solution?
CM = n / V
n = m / M
M (BeCl2) = 79.92 g/mol
n (BeCl2) = CM x V = 1.5 x 0.25 = 0.375 mol
m (BeCl2) = n x M = 0.375 x 79.92 = 29.97 g
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