Question

The freezing point of pure liquid  is . If the same amount of non-electrolyte non-volatile impurity is added in  and ideal solution is formed on cooling at 200 K, only 30%(by mass) liquid  is present and at 201 K, 60% (by mass) liquid  is present. The value of  is:​

Answer 1

Answer:

T-200/T-201=Kf×60/Kf×30

T-200=2T-402

T=202

Explanation:

∆T=kf×m

m when temperature is 200 is 60

m when temperature is 201 is 30

Answer 2

Answer:

Aim:

To find the temperature (freezing point) of the pure liquid:

T = 202

Explanation:

Freezing and boiling point of any liquid would be moderated in case of adding impurities to the pure liquid.  

Let us consider water for an instance, the normal boiling point of water is 100 degree Celsius a freezing point is zero degree Celsius.  

When salt is added to the pure water the boiling point of the water will result in increasing and the freezing point will result in decreasing.

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