Molarity , normality and molality of the solution containing 22% of `Al_2(SO_4)_3{d=1.253g//mL}` by weight are

Options

A. `0.825 M, 48.3 N,0.825m`
B. `0.805 M, 4.83 N,0.825m`
C. `4.83M, M, 4.83 N,4.83m`
D. None

Answer

To solve the problem of finding the molarity, normality, and molality of a solution containing 22% by weight of \( Al_2(SO_4)_3 \) with a density of 1.253 g/mL, we can follow these steps:

### Step 1: Understand the meaning of 22% by weight
22% by weight means that in a 100 g solution, there are 22 g of solute (which is \( Al_2(SO_4)_3 \)) and 78 g of solvent.

### Step 2: Calculate the volume of the solution
We can use the density to calculate the volume of the solution. The formula for density is:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
Rearranging gives us:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Substituting the values:
\[ \text{Volume} = \frac{100 \, \text{g}}{1.253 \, \text{g/mL}} \approx 79.84 \, \text{mL} \approx 0.07984 \, \text{L} \]

### Step 3: Calculate the number of moles of solute
Next, we need to calculate the number of moles of \( Al_2(SO_4)_3 \). The molar mass of \( Al_2(SO_4)_3 \) is calculated as follows:
- Aluminum (Al): \( 2 \times 27 \, \text{g/mol} = 54 \, \text{g/mol} \)
- Sulfur (S): \( 3 \times 32 \, \text{g/mol} = 96 \, \text{g/mol} \)
- Oxygen (O): \( 12 \times 16 \, \text{g/mol} = 192 \, \text{g/mol} \)

Adding these together gives:
\[ \text{Molar mass of } Al_2(SO_4)_3 = 54 + 96 + 192 = 342 \, \text{g/mol} \]
Now, we can find the number of moles:
\[ \text{Moles of } Al_2(SO_4)_3 = \frac{22 \, \text{g}}{342 \, \text{g/mol}} \approx 0.0643 \, \text{mol} \]

### Step 4: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{0.0643 \, \text{mol}}{0.07984 \, \text{L}} \approx 0.805 \, \text{mol/L} \]

### Step 5: Calculate the normality of the solution
Normality (N) is related to molarity (M) by the equation:
\[ \text{Normality} = n \times \text{Molarity} \]
where \( n \) is the number of equivalents. For \( Al_2(SO_4)_3 \), it dissociates into 2 Al\(^{3+}\) ions and 3 SO\(_4^{2-}\) ions, giving a total charge of 6. Thus, \( n = 6 \):
\[ \text{Normality} = 6 \times 0.805 \approx 4.83 \, \text{N} \]

### Step 6: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. We have 78 g of solvent, which is 0.078 kg:
\[ \text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.0643 \, \text{mol}}{0.078 \, \text{kg}} \approx 0.824 \, \text{mol/kg} \]

### Final Results
- Molarity: \( 0.805 \, \text{mol/L} \)
- Normality: \( 4.83 \, \text{N} \)
- Molality: \( 0.824 \, \text{mol/kg} \)