Density of 1M solution of a non-electrolyte `C_(6)H_(12)O_(6)` is 1.18 g/mL . If `K_(f) (H_(2)O)` is `1.86^(@) mol^(-1)` kg , solution freezes at :

Options

A. `1.58^(@)C`
B. `-1.86^(@)C`
C. `-3.16^(@)C`
D. `1.86^(@)C`

Answer

To find the freezing point of a 1M solution of glucose (C₆H₁₂O₆) with a given density, we can follow these steps:

### Step 1: Calculate the mass of the solution
Given that the density of the solution is 1.18 g/mL and the volume of the solution is 1 L (1000 mL), we can calculate the mass of the solution.

\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.18 \, \text{g/mL} \times 1000 \, \text{mL} = 1180 \, \text{g} \]

### Step 2: Calculate the mass of the solute
Since we have a 1M solution, there is 1 mole of glucose in the solution. The molecular weight of glucose (C₆H₁₂O₆) can be calculated as follows:

\[ \text{Molecular weight of glucose} = (6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180 \, \text{g/mol} \]

Thus, the mass of the solute (glucose) is:

\[ \text{Mass of solute} = 1 \, \text{mole} \times 180 \, \text{g/mol} = 180 \, \text{g} \]

### Step 3: Calculate the mass of the solvent
The mass of the solvent can be found by subtracting the mass of the solute from the mass of the solution:

\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} = 1180 \, \text{g} - 180 \, \text{g} = 1000 \, \text{g} \]

### Step 4: Convert the mass of the solvent to kg
To calculate molality, we need the mass of the solvent in kilograms:

\[ \text{Mass of solvent in kg} = \frac{1000 \, \text{g}}{1000} = 1 \, \text{kg} \]

### Step 5: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Since we have 1 mole of glucose and 1 kg of solvent:

\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1 \, \text{mole}}{1 \, \text{kg}} = 1 \, \text{mol/kg} \]

### Step 6: Calculate the depression in freezing point
Using the formula for freezing point depression:

\[ \Delta T_f = K_f \times m \]

Where \( K_f \) for water is given as 1.86 °C kg/mol. Thus:

\[ \Delta T_f = 1.86 \, \text{°C kg/mol} \times 1 \, \text{mol/kg} = 1.86 \, \text{°C} \]

### Step 7: Calculate the freezing point of the solution
The freezing point of pure water is 0 °C. The freezing point of the solution can be calculated as follows:

\[ \text{Freezing point of solution} = \text{Freezing point of solvent} - \Delta T_f = 0 \, \text{°C} - 1.86 \, \text{°C} = -1.86 \, \text{°C} \]

### Final Answer
The solution freezes at -1.86 °C.

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