
Which of the following salts would have the greatest effect on the freezing point of solution if \[0.10\]mole of salt was dissolved in enough water to make \[100ml\]of the solution?
\[\begin{align}
& \text{A}\text{. NaCl} \\
& \text{B}\text{. KCl} \\
& \text{C}\text{. CaC}{{\text{l}}_{\text{2}}} \\
& \text{D}\text{. MgC}{{\text{l}}_{\text{2}}} \\
& \text{E}\text{. AlC}{{\text{l}}_{\text{3}}} \\
\end{align}\]
Answer
545.7k+ views
Hint:Freezing point of a pure solvent depends on the amount of solute getting dissolved in the solvent. When concentration remains the same for the solute highest change in freezing point is determined by van’t Hoff factor.
Complete step by step answer:
The above question belongs to the concept of solution and its colligative properties. Colligative properties are the properties of a solution which depends upon the concentration of the solute. For example, depression in freezing point, vapour pressure, elevation in boiling point, osmotic pressure. In solutions we determine two values of colligative properties, one is the observed value and other is the ideal value. The difference between these two values gives us the van’t Hoff factor. van’t Hoff factor can also be defined as the ratio of the actual concentration of particles to concentration of the substance calculated using mass.
In the above question we can’t use the concentration of salts to determine the greatest change in the freezing point as it is the same. Therefore, we will calculate the van’t Hoff factor for each salt and it is equal to no ions into which the dissociates.
In case of \[\text{NaCl}\] and \[\text{KCl}\] the value of van’t Hoff factor is equal to two, in case of \[\text{CaC}{{\text{l}}_{\text{2}}}\] and \[\text{MgC}{{\text{l}}_{\text{2}}}\] the value is three and in case of \[\text{AlC}{{\text{l}}_{\text{3}}}\] the value is four.
Greater is the value of van’t Hoff factor more will be depression in freezing point. Therefore, we can conclude that \[\text{AlC}{{\text{l}}_{\text{3}}}\] will cause the greatest effect in freezing point.
Hence, option (E) is the correct answer.
Note:
Depression in freezing point is a colligative property and depends upon the amount of solute present in the solution. Lowest depression is observed for the solutions which contain urea.
Complete step by step answer:
The above question belongs to the concept of solution and its colligative properties. Colligative properties are the properties of a solution which depends upon the concentration of the solute. For example, depression in freezing point, vapour pressure, elevation in boiling point, osmotic pressure. In solutions we determine two values of colligative properties, one is the observed value and other is the ideal value. The difference between these two values gives us the van’t Hoff factor. van’t Hoff factor can also be defined as the ratio of the actual concentration of particles to concentration of the substance calculated using mass.
In the above question we can’t use the concentration of salts to determine the greatest change in the freezing point as it is the same. Therefore, we will calculate the van’t Hoff factor for each salt and it is equal to no ions into which the dissociates.
In case of \[\text{NaCl}\] and \[\text{KCl}\] the value of van’t Hoff factor is equal to two, in case of \[\text{CaC}{{\text{l}}_{\text{2}}}\] and \[\text{MgC}{{\text{l}}_{\text{2}}}\] the value is three and in case of \[\text{AlC}{{\text{l}}_{\text{3}}}\] the value is four.
Greater is the value of van’t Hoff factor more will be depression in freezing point. Therefore, we can conclude that \[\text{AlC}{{\text{l}}_{\text{3}}}\] will cause the greatest effect in freezing point.
Hence, option (E) is the correct answer.
Note:
Depression in freezing point is a colligative property and depends upon the amount of solute present in the solution. Lowest depression is observed for the solutions which contain urea.
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