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        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?
3 months ago

Arun
23345 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 thatFreezing 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

3 months ago
Vikas TU
9775 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

3 months ago
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