Question

A stock solution is made by dissolving 51.6 g of Mg3N2 in enough water to make a 46.8 mL solution. 90.4 mL of this Mg3N2 solution are diluted to 200.0 mL. What is the final concentration of Mg3N2?

Answer 1

$$n(Mg_3N_2) = m/M = 51.6/100.95 = 0.27 mol$$

$$c_1 = n/V_1 = 0.27/0.0468 = 5.77M$$

$$n_1(Mg_3N_2) = c_1V_2 = 5.77*0.0904 = 0.52 mol$$

$$c_2 = 0.52/0.200 = 2.6M$$