Question

The freezing point of a diluted milk sample is found to be $-0.2^{\circ}C$, while it should have been $-0.5^{\circ}C$ for pure milk. How much water has been added to pure milk to make the diluted sample ?

• A. 2 cups of water to 3 cups of pure milk
• B. 1 cup of water to 3 cups of pure milk
• C. 3 cups of water to 2 cups of pure milk
• D. 1 cup of water to 2 cups of pure milk

Answer 1

Answer: (C)

Freezing point of milk $=-0.5^{\circ} C$
$\because \Delta T _{ f }=0.5^{\circ} C$
Freezing point of diluted milk $=-0.2^{\circ} C$
$\therefore \Delta T _{ f }=0.2^{\circ} C$
$\frac{\left(\Delta T_{t}\right)_{1}}{\left(\Delta T_{t}\right)_{1}}=\frac{0.5}{0.2}=\frac{k_{t} m}{k_{t} m}$
Both has same amount of solute.
Let y mole of solute
$\frac{0.5}{0.2}=\frac{y \times w_{2}}{w_{1} \times y} $
$\Rightarrow \frac{w_{2}}{w_{1}}=\frac{5}{2}$
$w_{2}=\frac{5}{2} w_{1}$