The freezing point of a diluted milk sample is found to be $-0.2^{\circ} \mathrm{C},$ while it should have…
The freezing point of a diluted milk sample is found to be $-0.2^{\circ} \mathrm{C},$ while it should have been $-0.5^{\circ} \mathrm{C}$ for pure milk. How much water has been added to pure milk to make the diluted sample?
1 cup of water to 2 cups of pure milk
3 cups of water to 2 cups of pure milk
1 cup of water to 3 cups of pure milk
2 cups of water to 3 cups of pure milk
Solution
Freezing point of diluted milk $=-0.2^{\circ} \mathrm{C}$
$
\Delta \mathrm{T}_{\mathrm{f}}^{*}=0.2^{\circ} \mathrm{C}
$
Free ring point ol pure milk $=-0.5^{\circ} \mathrm{C}$
$
\Delta \mathrm{T}_{\mathrm{f}}=0.5^{\circ} \mathrm{C}
$
$\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\Delta \mathrm{T}_{\mathrm{f}}^{\prime}}=\frac{\mathrm{K}_{\mathrm{f}} \times m}{\mathrm{~K}_{\mathrm{f}} \times m^{\prime}} ; m=\frac{\text { mole of solute }}{\text { mass of solvent }(\mathrm{kg})}$
Moles of solute are same in both samples.
$
\therefore \frac{0.5}{0.2}=\frac{\mathrm{W}^{n}}{\mathrm{~W}}
$
$\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{5}{2} ; \mathrm{W}^{\prime}=\frac{5}{2} \mathrm{~W}$
2 cups of pure milk is mixed with 3 cups of water to make
5 cups of diluted milk