The freezing point of 0.05 m solution of glucose in water is `(K1 = 1.86°C m^(-1)`)

Options

A. 0.093°C
B. -1.86°C
C. -0.093 °C
D. -0.93°C

Answer

To solve the problem of finding the freezing point of a 0.05 molal solution of glucose in water, we will follow these steps:

### Step 1: Understand the formula for freezing point depression
The freezing point depression (\( \Delta T_f \)) can be calculated using the formula:
\[ \Delta T_f = K_f \times m \]
where:
- \( K_f \) is the freezing point depression constant (given as 1.86 °C kg/mol for water),
- \( m \) is the molality of the solution (given as 0.05 molal).

### Step 2: Substitute the values into the formula
Now, we will substitute the values of \( K_f \) and \( m \) into the formula:
\[ \Delta T_f = 1.86 \, \text{°C kg/mol} \times 0.05 \, \text{mol/kg} \]

### Step 3: Calculate \( \Delta T_f \)
Now we perform the multiplication:
\[ \Delta T_f = 1.86 \times 0.05 = 0.093 \, \text{°C} \]

### Step 4: Determine the freezing point of the solution
The freezing point of the pure solvent (water) is 0 °C. To find the freezing point of the solution, we subtract the freezing point depression from the freezing point of the solvent:
\[ T_f = T_f^{\text{solvent}} - \Delta T_f \]
Substituting the values:
\[ T_f = 0 \, \text{°C} - 0.093 \, \text{°C} = -0.093 \, \text{°C} \]

### Step 5: Conclusion
Thus, the freezing point of the 0.05 molal solution of glucose in water is:
\[ \boxed{-0.093 \, \text{°C}} \]

---