Answer to Question #217310 in Chemistry for mmm

Question #217310

If 20.0mL of 0.400M NaCl is added to 80.0mL of 0.900M NaCl. What is the resulting concentration of the NaCl?

Expert's answer

CM = n /V

n = CM x V

n1 = 0.400 x 0.02 = 0.008 mol

n2 = 0.9 x 0.08 = 0.072 mol

CM3 = (0.008+0.072) / (0.02+0.08) = 0.8 M

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