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
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
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Always assume the density of pure water to be $1\text{ g mL}^{-1}$ (or $1\text{ kg L}^{-1}$) unless stated otherwise. This simplifies concentration conversions because the volume of water in litres directly equals its mass in kilograms.
Updated On: May 28, 2026
- $180\text{ gm}$
- $18\text{ gm}$
- $1.8\text{ gm}$
- $0.18\text{ gm}$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
We need to determine the mass of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) added to 1 litre of water given the ratio of freezing point depression to the cryoscopic constant ($\Delta T_f / K_f$).
Step 2: Key Formula or Approach:
The depression in freezing point ($\Delta T_f$) is given by:
\[ \Delta T_f = K_f \cdot m \]
Where:
- $K_f$ is the molal depression constant (cryoscopic constant).
- $m$ is the molality of the solution.
This can be rearranged as:
\[ m = \frac{\Delta T_f}{K_f} \]
Step 3: Detailed Explanation:
Given that:
\[ \frac{\Delta T_f}{K_f} = \frac{1}{1000} \]
Therefore, the molality of the glucose solution is:
\[ m = 0.001\text{ mol kg}^{-1} \]
Molality is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \]
For 1 litre of water (the solvent):
- Since the density of water is $\approx 1\text{ g mL}^{-1}$, the mass of 1 litre of water is $1000\text{ g} = 1\text{ kg}$.
Thus, the moles of glucose added are:
\[ \text{moles of glucose} = m \times \text{mass of solvent in kg} = 0.001\text{ mol kg}^{-1} \times 1\text{ kg} = 0.001\text{ mol} \]
The molar mass of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) is $180\text{ g mol}^{-1}$.
Now, calculate the mass ($w$) of glucose added:
\[ w = \text{moles} \times \text{molar mass} = 0.001\text{ mol} \times 180\text{ g mol}^{-1} = 0.18\text{ g} \]
Step 4: Final Answer:
The correct option is (D).
We need to determine the mass of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) added to 1 litre of water given the ratio of freezing point depression to the cryoscopic constant ($\Delta T_f / K_f$).
Step 2: Key Formula or Approach:
The depression in freezing point ($\Delta T_f$) is given by:
\[ \Delta T_f = K_f \cdot m \]
Where:
- $K_f$ is the molal depression constant (cryoscopic constant).
- $m$ is the molality of the solution.
This can be rearranged as:
\[ m = \frac{\Delta T_f}{K_f} \]
Step 3: Detailed Explanation:
Given that:
\[ \frac{\Delta T_f}{K_f} = \frac{1}{1000} \]
Therefore, the molality of the glucose solution is:
\[ m = 0.001\text{ mol kg}^{-1} \]
Molality is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \]
For 1 litre of water (the solvent):
- Since the density of water is $\approx 1\text{ g mL}^{-1}$, the mass of 1 litre of water is $1000\text{ g} = 1\text{ kg}$.
Thus, the moles of glucose added are:
\[ \text{moles of glucose} = m \times \text{mass of solvent in kg} = 0.001\text{ mol kg}^{-1} \times 1\text{ kg} = 0.001\text{ mol} \]
The molar mass of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) is $180\text{ g mol}^{-1}$.
Now, calculate the mass ($w$) of glucose added:
\[ w = \text{moles} \times \text{molar mass} = 0.001\text{ mol} \times 180\text{ g mol}^{-1} = 0.18\text{ g} \]
Step 4: Final Answer:
The correct option is (D).
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