Q7 / 155 · WBJEE 2026 · Chemistry
Glucose is added to $1$ litre of water to such an extent that $\Delta T_f/K_f$ equals to $\dfrac{1}{1000}$. The weight of glucose added is
- A$180$ gm
- B$18$ gm
- C$1.8$ gm
- Correct answerD$0.18$ gm
Solution
Using the formula for depression in freezing point:
$\Delta T_f = K_f \times m$
where $m$ is the molality of the solution.
Given $\dfrac{\Delta T_f}{K_f} = \dfrac{1}{1000}$
Therefore, $m = \dfrac{1}{1000}$ mol/kg.
Mass of $1$ litre of water is $1$ kg.
Molality $m = \dfrac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$
Moles of glucose $= \dfrac{1}{1000} \times 1 = 10^{-3}$ mol.
Molar mass of glucose $(C_6H_{12}O_6)$ is $180$ g/mol.
Weight of glucose added $= 10^{-3} \times 180 = 0.18$ gm.
Answer: $0.18$ gm
$\Delta T_f = K_f \times m$
where $m$ is the molality of the solution.
Given $\dfrac{\Delta T_f}{K_f} = \dfrac{1}{1000}$
Therefore, $m = \dfrac{1}{1000}$ mol/kg.
Mass of $1$ litre of water is $1$ kg.
Molality $m = \dfrac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$
Moles of glucose $= \dfrac{1}{1000} \times 1 = 10^{-3}$ mol.
Molar mass of glucose $(C_6H_{12}O_6)$ is $180$ g/mol.
Weight of glucose added $= 10^{-3} \times 180 = 0.18$ gm.
Answer: $0.18$ gm
Hence, the correct answer is option: 'D'.