Question
Glucose is added to 1 litre of water to such an extent that $\Delta T_f / K_f$ equals to $\frac{1}{1000}$. The weight of glucose added is
This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
0.18 gm
Calculating Glucose Weight from Freezing Point Depression
This problem involves calculating the mass of glucose added to water based on the change in freezing point properties. We can solve this using the principles of colligative properties.
Given Information:
- Volume of water = 1 litre (approximately 1 kg or 1000 g)
- Relationship: $\frac{\Delta T_f}{K_f} = \frac{1}{1000}$
- Solute: Glucose ($C_6H_{12}O_6$)
- Molar mass of glucose ($M$) = $180$ g/mol
- Glucose is a non-electrolyte, so the van't Hoff factor ($i$) = $1$
Relevant Formula:
The depression in freezing point ($\Delta T_f$) 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
- $K_f$ is the cryoscopic constant
- $m$ is the molality of the solution
Step-by-Step Calculation:
- Determine Molality: Rearrange the formula using the given information $\frac{\Delta T_f}{K_f} = \frac{1}{1000}$: $\frac{\Delta T_f}{K_f} = i \cdot m$ Since $i = 1$ for glucose, we have: $m = \frac{\Delta T_f}{K_f} = \frac{1}{1000}$ mol/kg.
- Relate Molality to Weight: Molality ($m$) is defined as moles of solute per kilogram of solvent. $m = \frac{\text{moles of glucose}}{\text{mass of water (kg)}}$ Moles of glucose can be calculated as $\frac{\text{weight of glucose}}{\text{molar mass of glucose}}$. So, $m = \frac{\text{weight of glucose} / M}{W_{solvent (kg)}}$
- Calculate Weight of Glucose: Substitute the known values: Mass of water ($W_{solvent}$) = 1 kg Molar mass of glucose ($M$) = 180 g/mol Molality ($m$) = $\frac{1}{1000}$ mol/kg $\frac{1}{1000} \text{ mol/kg} = \frac{\text{weight of glucose} (\text{g}) / 180 \text{ g/mol}}{1 \text{ kg}}$ Weight of glucose = $\frac{1}{1000} \times 180$ g Weight of glucose = $\frac{180}{1000}$ g Weight of glucose = $0.18$ g
Conclusion:
The weight of glucose added is $0.18$ g.
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