Question

The total molarity and normality of all the ions present in a solution containing 0.1 M of $CuSO_{4}$ and 0.1 M of $Al_{2}\left(\right.SO_{4}\left(\left.\right)_{3}$ is

• A. 0.2 M, 0.2 N
• B. 0.7 M, 1.6 N
• C. 0.8 M, 1.6 N
• D. 1.6 M, 1.6 N

Answer 1

Answer: (B)

mole of $Cu^{2 +}=0.1$ , mole of $SO_{4}^{2 -}=0.1$
mole of $Al^{3 +}=0.2$ , mole of $SO_{4}^{2 -}=0.3$
So total mole = 0.1 + 0.1 + 0.2 + 0.3 = 0.7
Also, eq. of $Cu^{2 +}=0.2$ , eq. of $SO_{4}^{2 -}=0.2$
Eq. of $Al^{3 +}=0.6,$ Eq. of $SO_{4}^{2 -}=0.6$
So total eq. $=0.2+0.2+0.6+0.6=1.6$ .