Question

How many mL of 4.0% Mg(NO3)2 solution would contain 1.2 g of magnesium nitrate?

Answer 1

4.0 % in mass per volume :

4.0 g ---------- 100 mL
1.2 g ----------  ( volume )

Volume =  1.2 x 100 / 4.0

Volume = 120 / 4.0

= 30 mL

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