The van't Hoff factor for ammonium chloride (NH4Cl) is approximately 1.77, and the van't Hoff factor forammonium sulfate((NH4)2SO4) is approximately 2.37.What are the values of the van't Hoff factors for these salts?The van't Hoff factor (i) represents the number ofparticles(ions or molecules) into which a compound dissociates or associates when it dissolves in a solvent. To calculate the van't Hoff factor for a given solute, you can use the formula:i = (ΔTf) / (Kf * molality)Where:- ΔTf is the depression of freezing point (given as the difference between the freezing point of the solvent and the freezing point of the solution).- Kf is the cryoscopic constant of the solvent (for water, Kf = 1.86°C kg/mol).- molality is the molality of the solution (moles of solute per kilogram of solvent).First, we need to calculate the molality (m) for each solution:For ammonium chloride (NH4Cl):- ΔTf for NH4Cl = -0.322°C- Kf for water = 1.86°C kg/mol- molality (m) = ΔTf / (Kf * i)Now, let's calculate the van't Hoff factor (i) for ammonium chloride:- m = 0.0956 m- ΔTf = -0.322°C- Kf = 1.86°C kg/moli (NH4Cl) = ΔTf / (Kf * m)i (NH4Cl) = (-0.322°C) / (1.86°C kg/mol * 0.0956 m)i (NH4Cl) ≈ 1.77Now, let's calculate the van't Hoff factor (i) for ammonium sulfate (NH4)2SO4:- m = 0.0358 m- ΔTf = -0.173°C- Kf = 1.86°C kg/moli ((NH4)2SO4) = ΔTf / (Kf * m)i ((NH4)2SO4) = (-0.173°C) / (1.86°C kg/mol * 0.0358 m)i ((NH4)2SO4) ≈ 2.37So, the van't Hoff factor for ammonium chloride (NH4Cl) is approximately 1.77, and the van't Hoff factor for ammonium sulfate ((NH4)2SO4) is approximately 2.37.Learn more aboutammonium sulfateatbrainly.com/question/36635547#SPJ1...

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