Answer to Question #192777 in General Chemistry for Nixon

General Chemistry

Question #192777

How many grams of beryllium chloride (BeCl₂) are needed to make 250mL of a 1.5M 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 0.25 = 0.375 mol

m (BeCl₂) = n x M = 0.375 x 79.92 = 29.97 g

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