Two students prepare two cyclohexane solutions having the same freezing point. S

Question

Two students prepare two cyclohexane solutions having the same freezing point. Student 1 uses 26.6 g of cyclohexane solvent and student 2 uses 24.1 g of cyclohexane solvent. Which student has the greater number of moles of solute? Show calculations. Two solutions are prepared using the same solute: Solution A: 0.27 g of the solute dissolves in 27.4 g of t - butanol Solution B: 0.23 g of the solute dissolves in 24.8 g of cyclohexane Which solution has the greatest freezing point change? Show calculations and explain.

Explanation / Answer

Depression in Freezing point = kf*m

Where m= molality=moles of solute/ kg of solvent

For 26.6 gm of cyclohexane solvent = 26.6/1000

Depression in Freezing point1 = kf*m1    (1)

For 24.1gm of cyclohexane =24.1/1000

Depression in Freezing point2= kf*m2   (2)

Eq.1/Eq.2 and noting that the freezing point is same

  Molaity, m1= m2 moles of solute in 1st *1000/26.6= moles of solute in 2 *1000/24.4

Moles of solute1/ Moles of solute2= 26.6/24.4=1.09

Moles of solute 1= 1.09* moles of solute 2

Moles of solute for 1st student is larger.

2.

Tf = Kf x m*i

where Tf is the change in freezing temperature, Kf is the freezing point depression constant, m is the molality of the solution (moles of solute per kilogram of solvent) and i is the van't Hoff factor.

Molecular weight of t-butanol : 74 and that of Cyclohexane= 84

For solution A

Mass of solvent= 27.4 gm =27.4/1000=0.0274 kg

Kf= 9.1 deg.c/kgmol , m= moles of solute/kg of solvent= 0.27/74/0.0274=0.133

Tf =9.1*0.133=1.211

For solution B

Mass of solvent= 24.8gm =24.8/1000=0.0248

Kf=20 deg.c/kgmol , m= moles of solute/ kg of solvent= (0.23/84)/0.0248=0.110

Tf= 0.110*20= 2.2

Cyclohexane will have more depression in freezing point

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