Why the freezing point depression of 0.4 m NaCl solution is nearly twice than that of 0.4 M glucose?

Answer the following question:  

Why the freezing point depression (ΔTf) of 0.4 M NaCl solution is nearly twice than that of 0.4 M glucose solution?

This is because Van’t Hoff factor for NaCl is 2 and for glucose, it is 1.

Concept: Relative Molecular Mass of Non-volatile Substances - Depression in Freezing Point

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Answer

Hint: “The freezing point depression is the difference in the freezing points of the solution from the pure solvent. This is true for any solute added to a solvent, the freezing point of the solution will be lower than the freezing point of the pure solvent”. Complete step by step answer:In the question it is given that there are two solutions.One solution is having 0.4 M Sodium chloride in water and the second solution is having 0.4 M glucose in the solution.Whenever sodium chloride dissolves in water it is going to split into two ions (means dissolves) as follows.\[NaCl\xrightarrow{{{H}_{2}}O}N{{a}^{+}}+C{{l}^{-}}\]But in case of glucose, it dissolves but is not going to split into two or more ions. Means stay as a single chemical in water.Both the solutions (0.4 M \[NaCl\] and 0.4 M glucose) have the same concentration of solutes but one solute dissociates into two ions (\[NaCl\]) and other solute glucose stay as a single component in the solution.So, the freezing point depression of 0.4 M \[NaCl\] solution (produces two ions) is nearly twice than that of 0.4 M glucose solution. This concept can be explained clearly by Van’t Hoff factor. “The Van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass”. According to Van't Hoff, Van’t Hoff factor for\[NaCl\] is 2 and for glucose, it is 1. Van’t Hoff factor is more means freezing point depression is more.So, the freezing point depression (\[\Delta {{T}_{f}}\]) of 0.4 M \[NaCl\] solution is nearly twice that of 0.4 M Glucose solution. Note: Don't be confused there is no relation between concentration of the solute to the freezing point depression of the solution. Because freezing point depression is directly proportional to the number of ions produced by the solute not on the concentration of the solute.

Vapour pressure of pure water at 298 K is 23.8 mm. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Vapoure pressure of pure water (solvent) at 298 K, p0 = 23.8 mmVapour pressure of solution, p = ?Mass of solvent ,W = 850 gMass of solute,M = 50 g

Mol. mass of water (H2O), M = 18 g mol–1Mol.mass of urea NH2 CO NH2

= 14 + 2 + 12 + 16 + 14 + 2

According to Raoult's law, p0-pp0=ωMWm

                     p=p0-w×Mm×W×p°

                   p=23.8-50×1860×850

                      =23.8-0.017=23.78

Answer

Verified

Hint: “The freezing point depression is the difference in the freezing points of the solution from the pure solvent. This is true for any solute added to a solvent, the freezing point of the solution will be lower than the freezing point of the pure solvent”. Complete step by step answer:In the question it is given that there are two solutions.One solution is having 0.4 M Sodium chloride in water and the second solution is having 0.4 M glucose in the solution.Whenever sodium chloride dissolves in water it is going to split into two ions (means dissolves) as follows.\[NaCl\xrightarrow{{{H}_{2}}O}N{{a}^{+}}+C{{l}^{-}}\]But in case of glucose, it dissolves but is not going to split into two or more ions. Means stay as a single chemical in water.Both the solutions (0.4 M \[NaCl\] and 0.4 M glucose) have the same concentration of solutes but one solute dissociates into two ions (\[NaCl\]) and other solute glucose stay as a single component in the solution.So, the freezing point depression of 0.4 M \[NaCl\] solution (produces two ions) is nearly twice than that of 0.4 M glucose solution. This concept can be explained clearly by Van’t Hoff factor. “The Van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass”. According to Van't Hoff, Van’t Hoff factor for\[NaCl\] is 2 and for glucose, it is 1. Van’t Hoff factor is more means freezing point depression is more.So, the freezing point depression (\[\Delta {{T}_{f}}\]) of 0.4 M \[NaCl\] solution is nearly twice that of 0.4 M Glucose solution. Note: Don't be confused there is no relation between concentration of the solute to the freezing point depression of the solution. Because freezing point depression is directly proportional to the number of ions produced by the solute not on the concentration of the solute.

Answer the following question:  

Why the freezing point depression (ΔTf) of 0.4 M NaCl solution is nearly twice than that of 0.4 M glucose solution?

This is because Van’t Hoff factor for NaCl is 2 and for glucose, it is 1.

Concept: Relative Molecular Mass of Non-volatile Substances - Depression in Freezing Point

  Is there an error in this question or solution?