-2
$\begingroup$

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>

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – MaxW Oct 15 '18 at 4:38

Not the answer you're looking for? Browse other questions tagged or ask your own question.