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
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),...
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