Question

In comparison to a 0.01M solution of sucrose, the depression in freezing point of a 0.01M CaCl2 solution is _____________.

Answer 1

Answer:

about three times

Explanation:

Here, we see that molality and Kf will be same as solvent is water.

So, depression in freezing point = i × m × Kf 

Now, i = vant hoff factor

It is calculated as follows:

i = Total number of ions after dissociation/association/Total number of ions before dissociation/association

i= 3 for MgCl2 as it dissociates into 3 ions

i = 1 for glucose as it does not undergo any association/dissociation.

Hence, greater the i greater will be depression. So, depression in freezing point of MgCl2 is three times.

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Answer 2

Answer:

ABOUT THREE TIMES

Explanation:

0.01 M solution of glucose does not ionize while 0.01 M MgCl2 solution furnishes 3 ions (Mg2+ + 2Cl–) in the solution, hence the value of colligative property for MgCl2 solution is about 3 times.

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