Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points from highest to lowest. CoCl3, NH4Cl, Li2SO4
The order of increasing freezing points is Explanation: The expression of depression in freezing point is given as: where, = Depression in freezing point i = Van’t Hoff factor = freezing point constant m = molality Let the molarity of the given solutions is ‘1 m’ For the given options: Option 1: 1 m Value of i = 4 So, molal concentration will be = Option 2: 1 m Value of i = 3 So, molal concentration will be = Option 3: 1 m Value of i = 2 So, molal concentration will be = As, the molal concentration of is the highest, so its freezing point will be the highest. Hence, the order of increasing freezing points follows:
By using the formula of depression in freezing, it can easily identify the rank of the given compounds. where, = freezing point depression i = van ‘t Hoff factor (number of ions per individual molecule of solute) m = molality of the solute = molal freezing point According to question, if concentration (m) and is same, then only van ‘t Hoff factor will change. For , i = 2 For , i = 4 For , i= 3 Now, the solute with the largest i value results in low freezing point. The highest value of i is 4 (), thus has lowest freezing point. And, has highest freezing point. The order of freezing point is :
CoCl₃ > Li₂SO₄ > NH₄I. Explanation: Adding solute to water causes depression of the boiling point.The depression in freezing point (ΔTf) can be calculated using the relation: ΔTf = i.Kf.m, where, ΔTf is the depression in freezing point. i is the van ‘t Hoff factor. 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. For most non-electrolytes dissolved in water, the van ‘t Hoff factor is essentially 1. Kf is the molal depression constant of water. m is the molality of the solution. (1) Li₂SO₄: i for Li₂SO₄ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3. ∴ ΔTb for (Li₂SO₄) = i.Kb.m = (3)(Kf)(m) = 3(Kf)(m). (2) NH₄I: i for NH₄I = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2. ∴ ΔTb for (NH₄I) = i.Kb.m = (2)(Kf)(m) = 2(Kf)(m). (3) CoCl₃: i for CoCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4. ∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(m) = 4(Kf)(m). So, the ranking of the freezing point from the highest to the lowest is: CoCl₃ > Li₂SO₄ > NH₄I.
CoCl₃ > Li₂SO₄ > NH₄I. Explanation: Adding solute to water causes depression of the boiling point.The elevation in boiling point (ΔTf) can be calculated using the relation: ΔTf = i.Kf.m, where, ΔTf is the depression in freezing point. i is the van ‘t Hoff factor. 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. For most non-electrolytes dissolved in water, the van ‘t Hoff factor is essentially 1. Kf is the molal depression constant of water. m is the molality of the solution. (1) Li₂SO₄: i for Li₂SO₄ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3. ∴ ΔTb for (Li₂SO₄) = i.Kb.m = (3)(Kf)(m) = 3(Kf)(m). (2) NH₄I: i for NH₄I = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2. ∴ ΔTb for (NH₄I) = i.Kb.m = (2)(Kf)(m) = 2(Kf)(m). (3) CoCl₃: i for CoCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4. ∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(m) = 4(Kf)(m). So, the ranking of the freezing point from the highest to the lowest is: CoCl₃ > Li₂SO₄ > NH₄I.
The freezing order from the largest to the smallest is:
NH4Cl> K2SO4> COBr3
Further explanation For electrolyte solutions: ΔTb = Kb.m. i ΔTf = Kf.m. i Kb = molal boiling point increase Kf = molal freezing point constant m = molal solution i = van’t Hoff factor i = 1 + (n-1) α The formula above shows that the freezing point depends on •molal value •degree of ionization/dissociation •number of ions in solution In the statement of the matter, it is known that there are solutions which have the same concentration and are completely dissociated (α = 1) So the value of the freezing point drop depends only on the number of ions in the solution The more ions produced, the greater the value of the freezing point (the freezing point is smaller) There are several solutions •1. NH4Cl ionization: NH4Cl —> NH4+ + Cl- There are 2 ions in the solution •2. COBr3 ionization: COBr3 —> CO3++ 3Br- There are 4 ions in the solution •3. ionization K2SO4 K2SO4 —> 2K ++ SO42- There are 3 ions in the solution So the order of freezing from the largest to the smallest is: NH4Cl> K2SO4> COBr3 Learn more The freezing point of a solution Keywords: freezing point, properties, van’t Hoff factor
Answer 6
Assuming equal concentrations and complete dissociation, the aqueous solutions with the highest freezing point is NA3PO4 floowed by K2SO4, and then NH4I. This is due tot he number of ions being dissociated in the solution that will help in depressing the freezing point of a solution.
Answer 7
Justification: 1) The depression of the freezing point of a solution is a colligative property, which means that it depends on the number of particles of solute dissolved. 2) The formula for the depression of freezing point is: ΔTf = i * Kf * m Where i is the van’t Hoof factor which accounts for the dissociation of the solute. Kf is the freezing molal constant and only depends on the solvent m is the molality (molal concentration). 3) Since, you are assuming equal concentrations and complete dissociation of the given solutes, the solute with more ions in the molecular formula will result in the solution with higher depression of the freezing point (lower freezing point). 4) These are the dissociations of the given solutes: a) NH4 Cl (s) –> NH4(+)(aq) + Cl(-) (aq) => 1 mol –> 2 moles b) Co Br3 (s) –> Co(3+) (aq) + 3Br(-)(aq) => 1 mol –> 4 moles c) K2SO4 (s) –> 2K(+) (aq) + SO4 (2-) (aq) => 1 mol –> 3 moles 5) So, the rank of solutions by their freezing points is: CoBr3 < K2SO4 < NH4 Cl
meaning that NaI has the highest freezing point, and Co Br3 has the lowest freezing point. Explanation: The freezing point depression is a colligative property. That means that it depends on the number of solute particles dissolved. The formula to calculate the freezing point depression of a solution of a non volatile solute is: ΔTf = i * Kf * m Where kf is a constant, m is the molality and i is the van’t Hoff factor. Molality, which is number of moles per kg of solvent, counts for the number of moles dissolved and the van’t Hoff factor multipllies according for molecules that dissociate. The higher the number of molecules that dissociate, the higher the van’t Hoff, the greater the freezing point depression and the lower the freezing point. As the question states that you assume equal concentrations (molality) and complete dissociation you just must find the number of ions generated by each solute, in this way: NH4 I → NH4(+) + I(-) => 2 ions Co Br3 → Co(+) + 3 Br(-) => 4 ions Na2SO4 → 2Na(+) + SO4(2-) => 3 ions. So, Co Br3 is the solute that generate more particles and that solution will exhibit the lowest freezing point among the options given, Na2SO4 is next and the NaI is the third. Ordering the freezing point from higher to lower the rank is NaI > Na2SO4 > CoBr3, which is the answer given.
When you have solutes that are ionic compounds they dissociate in water into ions, then the compound that dissociates more ions will produce more particles and will decrease the freezing point the most. Given theses aqueous solutions Na2 CO3, Co Cl3, and Li NO3 you can predict the order of the freezing points. First, write the dissociation equations> Na2CO3 -> 2Na(+) + CO3 (2-) These are 3 ions: two of Na(+) and one of CO3(2-) The number inside parenthesis are number of charge not number of molecules. Co Cl3 -> Co(3+) + 3 Cl (1-) Those are 4 ions: one of Co (+) and three of Cl (-) Li NO3 -> Li (+) + NO3 (-) those are two ions: one of Li (+) and one of NO3(-) Then the ionic compound that dissociates into more ions give the solution with lower freezing point, and these is the rank from higher to lower freezing point: Li NO3 > Na2 CO3 > Co Cl3.
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Freezing points:CoCl₃ < K₂SO₄ < NH₄I
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