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>
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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
answered Oct 15, 2018 at 0:22
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$\begingroup$ 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? $\endgroup$– MaxWCommented Oct 15, 2018 at 4:38