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

Options

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

Answer

To find the freezing point of a 0.05 m solution of glucose in water, we can follow these steps:

### Step 1: Identify the given values
- **Molality (m)** of the glucose solution = 0.05 m
- **Freezing point depression constant (Kf)** for water = 1.86 °C/m

### Step 2: Use the formula for freezing point depression
The formula for the depression in freezing point (ΔTf) is given by:
\[ \Delta Tf = K_f \times m \]

### Step 3: Substitute the values into the formula
Now, substituting the known values into the formula:
\[ \Delta Tf = 1.86 \, °C/m \times 0.05 \, m \]

### Step 4: Calculate ΔTf
Now, perform the multiplication:
\[ \Delta Tf = 1.86 \times 0.05 = 0.093 \, °C \]

### Step 5: Determine the new freezing point
The freezing point of pure water is 0 °C. The new freezing point (Tf) can be calculated using the formula:
\[ Tf = Tf_0 - \Delta Tf \]
where \(Tf_0\) is the freezing point of pure water (0 °C).

Substituting the values:
\[ Tf = 0 °C - 0.093 °C = -0.093 °C \]

### Final Answer
The freezing point of the 0.05 m solution of glucose in water is **-0.093 °C**.

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