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Question: In comparison to a \(0.01M\) solution of glucose, the depression in freezing point of a \(0.01M\) \(...

In comparison to a 0.01M0.01M solution of glucose, the depression in freezing point of a 0.01M0.01M MgCl2MgC{l_2} solution is?
A) The same.
B) About twice.
C) About three times.
D) About six times.

Explanation

Solution

We all know depression in melting point is directly associated with van't Hoff factor (i)\left( i \right) consistent with which greater the worth of van't Hoff factor greater are going to be Depression in melting point.

Complete step by step answer:
We can calculate the melting point depression using the formula,
ΔTf=i×Kf×molality\Delta Tf = i \times Kf \times molality
Where , the melting point depression is denoted as ΔTf\Delta Tf, KfKf is that the melting point depression constant, and that ii is that the van ‘t Hoff factor.
The van't Hoff factor (i)\left( i \right) is the total number of ions after dissociation or association or the total number of ions before dissociation or association.
Given,
The molality of glucose is 0.01M0.01M.
The molality of MgCl2MgC{l_2} is0.01M0.01M.
iglucose=1{i_{glu\cos e}} = 1
ΔTfglucose=1×Kf×0.01\Delta T{f_{glu\cos e}} = 1 \times Kf \times 0.01
The dissociation equation of magnesium chloride is,
MgCl2Mg2++2ClMgC{l_2}\xrightarrow{{}}M{g^{2 + }} + 2C{l^ - }
i=3i = 3
ΔTfMgCl2=3×Kf×0.01\Delta T{f_{MgC{l_2}}} = 3 \times Kf \times 0.01
ΔTfMgCl2=3×ΔTfglucose\Delta T{f_{MgC{l_2}}} = 3 \times \Delta T{f_{glu\cos e}}
For MgCl2MgC{l_2} the van't Hoff factor (i)\left( i \right) is three as it dissociates into three ions and for glucose the van't Hoff factor (i)\left( i \right)is one as it does not undergo any association or dissociation. Thus, greater the van't Hoff factor (i)\left( i \right) greater will be depression. So, depression in the freezing point of MgCl2MgC{l_2} is three times.

Therefore, the option C is correct.

Note:
As we know that the freezing point depression is the phenomena that describes why adding a solute to a solvent leads to the lowering of the melting point of the solvent. When a substance starts to freeze, the molecules hamper thanks to the decreases in temperature, and therefore the intermolecular forces start to require over. The molecules will then arrange themselves during a pattern, and thus become a solid. For instance, as water is cooled to the melting point, its molecules become slower and hydrogen bonds begin to stick more, eventually creating a solid. If salt is added to the water, the Na+N{a^ + } and ClC{l^-} ions attract to the water molecules and interfere with the formation of the massive network solid referred to as ice. So as to realize a solid, the answer must be cooled to a good lower temperature.