Step 1: Understanding the Concept:
Colligative properties like the depression in freezing point ($\Delta T_f$) are directly proportional to the molality ($m$) of the solution.
The relationship is given by the formula $\Delta T_f = K_f \cdot m \cdot i$, where $K_f$ is the cryoscopic constant and $i$ is the van't Hoff factor.
Glucose ($C_6H_{12}O_6$) is a non-electrolyte, so it does not dissociate or associate in water, meaning $i = 1$.
Step 2: Key Formula or Approach:
1. Use the ratio $\Delta T_f / K_f$ to find the molality $m$.
2. Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$.
3. Weight of glucose = moles $\times$ molecular weight.
Step 3: Detailed Explanation:
From the formula $\Delta T_f = K_f \cdot m$, we can write:
$m = \Delta T_f / K_f$.
Given that $\Delta T_f / K_f = 1/1000$, the molality of the solution is $0.001 \text{ mol/kg}$.
The volume of water is $1$ litre.
Since the density of water is approximately $1 \text{ g/mL}$, the mass of $1$ litre of water is $1000 \text{ g}$ or $1 \text{ kg}$.
Now, calculate the moles of glucose present:
Moles = molality $\times$ mass of solvent (kg)
Moles = $0.001 \text{ mol/kg} \times 1 \text{ kg} = 0.001 \text{ moles}$.
The molecular formula of glucose is $C_6H_{12}O_6$.
Molecular weight $= (12 \times 6) + (1 \times 12) + (16 \times 6) = 72 + 12 + 96 = 180 \text{ g/mol}$.
Weight of glucose added = moles $\times$ molecular weight
Weight $= 0.001 \text{ mol} \times 180 \text{ g/mol} = 0.18 \text{ g}$.
Step 4: Final Answer:
The weight of glucose added to the water is 0.18 gm. The correct option is (D).
Step 5: Final Verification:
Units: $0.001 \text{ mol} \times 180 \text{ g/mol}$ results in $0.18 \text{ g}$. All steps are mathematically sound.