Answer to Question #149049 in General Chemistry for Tisha

General Chemistry

Question #149049

How man grams of magnesium cyanide are needed to make 275 mL of a 0.075 M solution?

Expert's answer

n(Mg(CN)₂) = V*c = (0.275 L)*(0.075 M) = 0.020625 mol

m(Mg(CN)2) = Mr(Mg(CN)2)*n(Mg(CN)2) = (76.3398 g/mol)*(0.020625 mol) = 1.5745 g.

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