Answer :
Answer:
option c, 0.1 M [tex]AlCl_3[/tex]
Explanation:
Addition of non-volatile solute to a solvent decreases its vapour pressure which results in decrease in melting point.
Decrease in melting point is known as depression in freezing point. the depression in freezing point is related with molality and no. of ions as follows:
[tex]\Delta T_f = imK_f[/tex]
Where, i is von't Hoff factor, m is molatilty and ΔTf is depression in freezing point.
As, in the given case concentration of all the solution is same, therefore, depression in freezing point will depend upon the no. of ions produced by the ionization of the salts in the aqueous solution.
In case of NaCl, the no. of ions produced will be 2.
Therefore, value of i will be 2
In case of [tex]MgCl_2[/tex], the no. of ions produced will be 3.
Therefore, value of i will be 3
In case of [tex]AlCl_3[/tex], the no. of ions produced will be 4.
Therefore, value of i will be 4.
More, the value of van't Hoff factor, more will be depression in freezing point.
Therefore, assuming the given solution to be aqueous, the solution expected to have lowest freezing point is 0.1 M [tex]AlCl_3[/tex].
The compound that has the lowest melting point is [tex]\bold{0.1 M\; AlCl_3}[/tex]
The correct option is [tex]\bold{0.1 M\; AlCl_3}[/tex] (C).
What is the freezing point?
- The temperature at which a liquid turns into a solid is known as the freezing point.
- Increased pressure raises the freezing point in the same way that it raises the melting point.
- The freezing point depression is T = KFm, where KF is the molal freezing point depression constant and m is the molality of the solute.
- Mol solute = (m) x (kg solvent), where kg of solvent is the mass of the solvent (lauric acid) in the mixture.
Thus, the correct option is (c) [tex]\bold{0.1 M\; AlCl_3}[/tex].
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