A 1.24 M solution of KI has a density of 1.15 g/cm3. Assume 1.00 liter of 1.24 M
ID: 304392 • Letter: A
Question
A 1.24 M solution of KI has a density of 1.15 g/cm3. Assume 1.00 liter of 1.24 M KI solution. The mass of 1.00 L of solution is grams. The number of grams of Kl in the 1 liter is grams. The number of grams of water in the 1 L is grams. The molality of the Kl solution is molal. The molal freezing point depression constant for water in kg/molal. The van't Hoff factor for complete dissociation is complete dissociation of Kl the freezing point depression is the actual freezing point depression is -4.46 deg C, the van't Hoff factor value is deg C . For deg C. If and the percent of dissociation is (2 S.F.) Molality is moles of solute per of solvent.Explanation / Answer
Basis: 1 Liter of KI solution, its density is 1.15 g/cc
1 liter =1000cc, hence mass of 1Liter solution is= volume in cc* density = 1000*1.15= 1150 gm
moles of KI in solution = molarity* volume in liters= 1.24*1= 1.24 moles
molar mass of KI= 166, mass of KI in solution = moles* molar mass= 1.24*166 =206 gm
mass of water in solution = total mass of solution- mass of KI = 1150-206= 944 gm
1000gm= 1kg, mass of water in kg= 944/1000=0.944 kg
molality= moles of solute (KI)/ kg of water = 1.24/0.944= 1.313 m
the molal freezing point depression constant for water, Kf= 1.86 deg.c/molal
when KI dissociated, it produces K+ and I- ions. Van't Hoff factor is the no of ions in solution. hence i = Van;t Hoff factor for complete dissocaiton = 2
Freezing point depression = i*molality* Kf =2*1.86*1.313 =4.88 deg.c
Freezing point = Freezing point of water- Freezing point depression = 0-4.88=-4.88
Given Freezing point = -4.46 deg.c, Freezing point depression = 0-(-4.46)= 4.46
i= 4.46/(1.313*1.86)= 1.83
% of dissociation = 100* actual i/ theoretical I= 100*1.83/2= 91.5%
molality is moles of solute per kg of solvent.
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