`30mL` of `CH_(3)OH (d = 0.780 g cm^(-3))` and `70 mL` of `H_(2)O (d = 0.9984 g cm^(-3))` are mixed at `25^(@)C` to form a solution of density `0.9575 g cm^(-3)`. Calculate the freezing point of the solution. `K_(f)(H_(2)O)` is `1.86 kg mol^(-1)K`. Also calculate its molarity.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the mass of CH₃OH (methanol)
Given:
- Volume of CH₃OH = 30 mL
- Density of CH₃OH = 0.780 g/cm³

Using the formula for mass:
\[ \text{Mass} = \text{Volume} \times \text{Density} \]

Calculating the mass:
\[ \text{Mass of CH₃OH} = 30 \, \text{mL} \times 0.780 \, \text{g/cm}^3 = 23.94 \, \text{g} \]

### Step 2: Calculate the mass of H₂O (water)
Given:
- Volume of H₂O = 70 mL
- Density of H₂O = 0.9984 g/cm³

Using the same formula:
\[ \text{Mass of H₂O} = 70 \, \text{mL} \times 0.9984 \, \text{g/cm}^3 = 69.89 \, \text{g} \]

### Step 3: Calculate the total mass of the solution
Total mass of the solution is the sum of the masses of CH₃OH and H₂O:
\[ \text{Total mass of solution} = \text{Mass of CH₃OH} + \text{Mass of H₂O} \]
\[ \text{Total mass of solution} = 23.94 \, \text{g} + 69.89 \, \text{g} = 93.83 \, \text{g} \]

### Step 4: Calculate the volume of the solution
Given the density of the solution:
- Density of solution = 0.9575 g/cm³

Using the formula for volume:
\[ \text{Volume of solution} = \frac{\text{Total mass}}{\text{Density}} \]
\[ \text{Volume of solution} = \frac{93.83 \, \text{g}}{0.9575 \, \text{g/cm}^3} \approx 98.06 \, \text{mL} \]

### Step 5: Calculate the molality of the solution
First, we need to find the moles of CH₃OH:
- Molar mass of CH₃OH = 32 g/mol

Calculating moles of CH₃OH:
\[ \text{Moles of CH₃OH} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{23.94 \, \text{g}}{32 \, \text{g/mol}} \approx 0.748 \, \text{mol} \]

Now calculate the weight of the solvent (H₂O) in kg:
\[ \text{Weight of H₂O} = 69.89 \, \text{g} = 0.06989 \, \text{kg} \]

Now calculate molality (m):
\[ m = \frac{\text{Moles of solute}}{\text{Weight of solvent in kg}} = \frac{0.748 \, \text{mol}}{0.06989 \, \text{kg}} \approx 10.69 \, \text{mol/kg} \]

### Step 6: Calculate the freezing point depression
Using the formula:
\[ \Delta T_f = K_f \times m \]
Where \( K_f \) for water = 1.86 °C kg/mol.

Calculating the freezing point depression:
\[ \Delta T_f = 1.86 \, \text{°C kg/mol} \times 10.69 \, \text{mol/kg} \approx 19.91 \, \text{°C} \]

### Step 7: Calculate the freezing point of the solution
The freezing point of pure water is 0 °C, so:
\[ T_f = T_{f, \text{pure}} - \Delta T_f \]
\[ T_f = 0 \, \text{°C} - 19.91 \, \text{°C} \approx -19.91 \, \text{°C} \]

### Step 8: Calculate the molarity of the solution
Molarity (M) is defined as:
\[ M = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} \]

Converting volume from mL to L:
\[ \text{Volume of solution} = 98.06 \, \text{mL} = 0.09806 \, \text{L} \]

Now calculating molarity:
\[ M = \frac{0.748 \, \text{mol}}{0.09806 \, \text{L}} \approx 7.63 \, \text{mol/L} \]

### Final Answers:
- Freezing point of the solution: **-19.91 °C**
- Molarity of the solution: **7.63 mol/L**