Question

The freezing point of a diluted milk sample is found to be $-{0.2}^{o}C$, while it should have been $-{0.5}^{o}C$ for pure milk. How much water has been added to 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. $3$ cup of water to $2$ cups of pure milk
• D. $1$ cup of water to $2$ cups of pure milk

Answer 1

The correct option is C $3$ cup of water to $2$ cups of pure milk

We have,

$T_f=-{0.5}^{o}C$ (For milk) and $T_f=-{0.2}^{o}C$ (For diluted solution)

$\therefore \Delta T_f={0.5}^{o}C$ (For milk) and $\Delta T-f={0.2}^{o}C$ (For diluted solution)

Now, we know,

$\Delta T_f=K_f\times m$

where $m=molality$

We can have,

$\frac{(\Delta T_f{)}_1}{(\Delta T_f{)}_2}=\frac{K_f\times x\times 1000w_2}{w_1\times K_f\times x\times 1000}$

\dfrac{0.5}{0.2}=\dfrac{w-2}{w-1}$

$\therefore w_2=\frac{5}{2}{w}_1$

$\therefore W_{water}=\frac{5}{2}{w}_1-w_1=\frac{2}{3}{w}_1$

Thus, we required 3 cup of water and 2 cup of milk