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(SO_4)_3` is

Options

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

To find the total molarity and normality of all the ions present in a solution containing 0.1 M of `CuSO4` and 0.1 M of `Al2(SO4)3`, we will follow these steps:

### Step 1: Determine the dissociation of `CuSO4`
`CuSO4` dissociates into:
- 1 mole of `Cu^2+`
- 1 mole of `SO4^2-`

Since the concentration of `CuSO4` is 0.1 M, the molarity of the ions will be:
- Molarity of `Cu^2+` = 0.1 M
- Molarity of `SO4^2-` = 0.1 M

### Step 2: Determine the dissociation of `Al2(SO4)3`
`Al2(SO4)3` dissociates into:
- 2 moles of `Al^3+`
- 3 moles of `SO4^2-`

Since the concentration of `Al2(SO4)3` is 0.1 M, the molarity of the ions will be:
- Molarity of `Al^3+` = 2 × 0.1 M = 0.2 M
- Molarity of `SO4^2-` = 3 × 0.1 M = 0.3 M

### Step 3: Calculate the total molarity of all ions
Now, we can sum up the molarities of all the ions:
- Molarity of `Cu^2+` = 0.1 M
- Molarity of `SO4^2-` from `CuSO4` = 0.1 M
- Molarity of `Al^3+` = 0.2 M
- Molarity of `SO4^2-` from `Al2(SO4)3` = 0.3 M

Total molarity of ions (MT):
\[ MT = [Cu^{2+}] + [SO_4^{2-}] + [Al^{3+}] + [SO_4^{2-}] \]
\[ MT = 0.1 + 0.1 + 0.2 + 0.3 = 0.7 \, M \]

### Step 4: Calculate the normality of the solution
Normality (N) is calculated using the formula:
\[ N = M \times n \]
where \( n \) is the valency factor.

1. For `CuSO4`:
- Valency factor (n) = 2 (for `Cu^2+`)
- Normality = 0.1 M × 2 = 0.2 N

2. For `Al2(SO4)3`:
- Valency factor (n) = 6 (2 × 3 for `Al^3+`)
- Normality = 0.1 M × 6 = 0.6 N

### Step 5: Calculate total normality
Now, we can sum up the normalities of all the ions:
- Normality of `Cu^2+` = 0.2 N
- Normality of `SO4^2-` from `CuSO4` = 0.2 N
- Normality of `Al^3+` = 0.6 N
- Normality of `SO4^2-` from `Al2(SO4)3` = 0.6 N

Total normality (NT):
\[ NT = [Cu^{2+}] + [SO_4^{2-}] + [Al^{3+}] + [SO_4^{2-}] \]
\[ NT = 0.2 + 0.2 + 0.6 + 0.6 = 1.6 \, N \]

### Final Answer
- Total Molarity of ions = 0.7 M
- Total Normality of ions = 1.6 N