Calculate the final molarity from 2 solutions [closed]

How do I calculate the final molarity when 70.0 mL of 3.0 M sodium chloride solution is added to 30.0 mL of a 1.00 M solution of sodium chloride>

closed as off-topic by Mithoron, airhuff, Jon Custer, A.K., WaylanderOct 15 '18 at 9:12

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Molarity (M) = moles (n) / liters (L)

70 mL of 3M NaCl = .21 mol NaCl
30 ml of 1M NaCl = .03 mol NaCl

.21 mol + .03 mol = .24 mol NaCl 70 mL + 30 mL = 100 mL = .1 L

M = n/L = .24 / .1 = 2.4 M

• The hidden assumption here is that the volumes of the solutions are additive. It is not strictly true. However for the solutions involved, probably close enough. // 2 significant figures or 3? Did the original problem statement specify the first solution as 3.0 M or 3.00 M? – MaxW Oct 15 '18 at 4:38