Hunderd gram of `Al(NHO_(3))_(3)` [molar mass `213 g mol^(-1)`] is dissolved in 1L of water at `20^(@)C`. The density of water at this temperature is `0.9982 cm^(-3)` and the density of resulting solution is `0.999 g cm^(-3)` . Calculate the molarity and molality of this solution .

Explanation

To solve the problem of calculating the molarity and molality of the solution formed by dissolving 100 g of `Al(NH4)3` in 1 L of water, we will follow these steps:

### Step 1: Calculate the number of moles of `Al(NH4)3`
To find the number of moles, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Given:
- Mass of `Al(NH4)3` = 100 g
- Molar mass of `Al(NH4)3` = 213 g/mol

Calculating the number of moles:

\[ \text{Number of moles} = \frac{100 \text{ g}}{213 \text{ g/mol}} \approx 0.469 \text{ moles} \]

### Step 2: Calculate the weight of water
We know the density of water at 20°C is 0.9982 g/cm³, and the volume of water is 1 L (which is equal to 1000 cm³).

Using the formula for density:

\[ \text{Density} = \frac{\text{mass}}{\text{volume}} \]

Rearranging gives us:

\[ \text{mass} = \text{Density} \times \text{volume} \]

Calculating the mass of water:

\[ \text{mass of water} = 0.9982 \text{ g/cm}^3 \times 1000 \text{ cm}^3 = 998.2 \text{ g} \]

### Step 3: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent (water in this case).

\[ \text{Molality} = \frac{\text{number of moles of solute}}{\text{mass of solvent (kg)}} \]

Converting the mass of water from grams to kilograms:

\[ \text{mass of water (kg)} = \frac{998.2 \text{ g}}{1000} = 0.9982 \text{ kg} \]

Now, substituting the values into the molality formula:

\[ \text{Molality} = \frac{0.469 \text{ moles}}{0.9982 \text{ kg}} \approx 0.469 \text{ mol/kg} \]

### Step 4: Calculate the volume of the solution
The density of the resulting solution is given as 0.999 g/cm³. The total mass of the solution is the sum of the mass of the solute and the solvent.

\[ \text{mass of solution} = \text{mass of solute} + \text{mass of solvent} = 100 \text{ g} + 998.2 \text{ g} = 1098.2 \text{ g} \]

Using the density to find the volume of the solution:

\[ \text{Density} = \frac{\text{mass}}{\text{volume}} \implies \text{volume} = \frac{\text{mass}}{\text{density}} \]

Calculating the volume:

\[ \text{Volume of solution} = \frac{1098.2 \text{ g}}{0.999 \text{ g/cm}^3} \approx 1099.3 \text{ cm}^3 \]

### Step 5: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute per liter of solution.

\[ \text{Molarity} = \frac{\text{number of moles of solute}}{\text{volume of solution (L)}} \]

Converting the volume from cm³ to liters:

\[ \text{Volume of solution (L)} = \frac{1099.3 \text{ cm}^3}{1000} = 1.0993 \text{ L} \]

Now substituting the values into the molarity formula:

\[ \text{Molarity} = \frac{0.469 \text{ moles}}{1.0993 \text{ L}} \approx 0.427 \text{ M} \]

### Final Results
- **Molarity**: 0.427 M
- **Molality**: 0.469 mol/kg