Question

5. If 75.0 mL of a 0.20 M solution of sodium nitrate (NaNO3) is mixed with 25.0 mL of 0.10 M barium nitrate (Ba(NO3)2), what is the molar concentration of nitrate in the resulting solution? A. 0.10 M D. 0.15 M B. 0.20 M E. 0.18 M C. 0.30 M

Answer 1

Concentration of mixture = (n1*C1*V1+ n2*C2*V2) / (V1+V2)
where C1 --> Concentration of 1 component
V1-->volume of 1 component
C2 --> Concentration of other component
V2-->volume of other component
n1 --> number of particle from 1 molecule of 1st component
n1 = 1 as 1 NaNO3 has 1 NO3- ion
n2 --> number of particle from 1 molecule of 2nd component
n2 = 2 as 1 Ba(NO3)2 has 2 NO3- ion

use:
C = (n1*C1*V1+ n2*C2*V2) / (V1+V2)
C = (1*0.2*75+2*0.1*25)/(75+25)
C = 0.20 M
Answer: B