Glucose is added to 1 litre water to an extent that `DeltaT_(f)//K_(f)` becomes equal to 1/1000, the mass of glucose added is

Options

A. 180 gm
B. 18 gm
C. 1.8 gm
D. 0.18 gm

Answer

To solve the problem, we need to determine the mass of glucose added to 1 liter of water such that the depression in freezing point (ΔTf) divided by the freezing point depression constant (Kf) equals 1/1000.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
We know that the depression in freezing point (ΔTf) is related to the molality (M) of the solution by the formula:
\[ \Delta T_f = i \cdot K_f \cdot M \]
where \(i\) is the van 't Hoff factor. For glucose, which is a non-electrolyte, \(i = 1\). Thus, the equation simplifies to:
\[ \Delta T_f = K_f \cdot M \]

2. **Given Information**:
We are given that:
\[ \frac{\Delta T_f}{K_f} = \frac{1}{1000} \]
This implies:
\[ M = \frac{1}{1000} \]

3. **Calculating Molality (M)**:
Molality (M) is defined as the number of moles of solute per kilogram of solvent. Since we have 1 liter of water, and the density of water is approximately 1 g/cm³, the mass of the solvent (water) is:
\[ 1 \text{ liter} = 1000 \text{ g} = 1 \text{ kg} \]
Therefore, we can express molality as:
\[ M = \frac{\text{number of moles of glucose}}{1 \text{ kg}} \]

4. **Finding the Number of Moles of Glucose**:
From the molality equation, we have:
\[ \frac{\text{number of moles of glucose}}{1 \text{ kg}} = \frac{1}{1000} \]
This means:
\[ \text{number of moles of glucose} = \frac{1}{1000} \text{ moles} \]

5. **Calculating the Mass of Glucose**:
The molar mass of glucose (C₆H₁₂O₆) is calculated as follows:
\[ \text{Molar mass} = (6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180 \text{ g/mol} \]
Now, we can find the mass of glucose using the number of moles:
\[ \text{mass of glucose} = \text{number of moles} \times \text{molar mass} = \frac{1}{1000} \text{ moles} \times 180 \text{ g/mol} = 0.18 \text{ g} \]

### Final Answer:
The mass of glucose added is **0.18 grams**.