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Grade: 12th pass
The freezing point of a diluted milk sample is found to be -0.2degreeC, while it should have been -0.5degreeC for pure milk. How much water has been added to pure milk to make the diluted sample? 
9 months ago

Answers : (2)

24735 Points

Depression in freezing point -





Freezing -

 Freezing  occurs when liquid solvent is in equilibrium with solid solvent. As non volatile solute decreases, the vapour pressure freezing point decreases.

-As we knoew that

Freezing point of milk = -0.5^{0}C\because \Delta T_{f}= 0.5^{0}C

Freezing point of milk (diluted)=-0.2^{0}C \because \Delta T_{f}= 0.2^{0}C

\frac{\left ( \Delta T_{f} \right )i}{\left ( \Delta T_{f} \right )ii}= \frac{0.5}{0.2}= \frac{K_{f}m}{K_{f}m}= \frac{x\left ( mole \right )\times weight\left ( 2 \right )}{weight_{\left ( 1 \right )\times }x\left ( mole \right )}

W_{2}= \frac{5}{2}W_{1}


3 cups of water into 2 cups of pure milk

9 months ago
Vikas TU
11684 Points
Dear student 
del Tf = Kf m 
m - molality 
molality is inversly proportional to mass of solvent 
del T /del Tf = m'/m 
Let x amount of water is added 
Initial mass = M 
mass of solvent = (M+x) 
del T / del Tf = (M+x)/M 
0.5/0.2 = (M+x)/M 
x = 1.5 M 
hence, M kg of pure milk 1.5M kg water should be added to dilute sample
9 months ago
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