Using data from the table, calculate the freezing and boiling points of each of

Question

Using data from the table, calculate the freezing and boiling points of each of the following solutions.

A) freezing point of 0.260 mol of naphthalene (C10H8) in 2.60 mol of chloroform

B) boiling point of 0.260 mol of naphthalene (C10H8) in 2.60 mol of chloroform

C) freezing point of 2.03 g KBr and 4.81 g glucose (C6H12O6) in 189 g of water

D) boiling point of 2.03 g KBr and 4.81 g glucose (C6H12O6) in 189 g of water

E) freezing point of 1.50 g NaCl in 0.250 kg of water

F) boiling point of 1.50 g NaCl in 0.250 kg of water

G) boiling point of 0.20 m glycerol (C3H8O3) in ethanol

H) freezing point of 0.20 m glycerol (C3H8O3) in ethanol

Question 1 TABLE 13.3 Molal Boiling-Point-Elevation and Freezing-Point-Depression Constants Normal Freezing Normal Boiling K60C/m) Point (°C Solvent Poin (C) 1.86 Water, H20 0.51 100.0 0.0 Benzene, C6H6 80.1 2.53 5.12 78.4 1.99 Ethanol, C2HOH 22 4.6 29.8 Carbon tetrachloride, CCI 76.8 5.02 -22.3 Chloroform, CHC, 3.63 4.68 61.2 63.5

Explanation / Answer

Freezing point= Freezing point of solvent- freezing point depression

Boiling point = Boiling point of solvent+ boiling point elevation

1.moles of chloroform = 2.6, molar mass of Chloroform (CHCl3) =3*35.5+12+1=119.5.

Mass of chloroform = 2.6*119.5 =310.7 gm = 0.3107 kg

molality = moles of solute/ kg of solvent = 0.260/0.3107=0.84

Freezing point depression = i*kf*m, i= 1 for napthalene kf= 4.68 deg.c/m and m= 0.84

Freezing point depresion = 1*4.68*0.84=3.93

Freezing point depresion = -63.5-3.93=-67.43 deg,c

2. Boiling point elevation = i*kb*m =1*3.63*0.84=3.05

Boilng point = 61.20+3.05 =64.25 deg.c

3. Freezing point of water in presnce of KBr,

moles of KBr in 2.03 gm = mass/molar mass = 2.03/119=0.0170 and moles of glucose = 4.81/180 =0.027

molality : KBr= 0.0170/0.189 kg water =0.089 and molality of glucose = 0.027/0.189=0.1429

Van;t Hoff factor for KBr= 2 and for glucose is =1

hence freezing point depression = 2*1.86*0.089+1*0.1429*1.86 =0.59

Freezing point =0-0.59= -0.59 deg.c

d) boiling point elevation = 2*0.51*0.089+1*0.1429*0.51=0.163

Boiling point= 100 +0.163= 100.163 deg.c

e) i=2 for NaCl, molalty of Nacl = moles/ kg solvent = 1.5/(58.5*0.25)=0.102564

Freezing point depression =i*kf*m =2*1.86*0.102564=0.38

Freezing point =0-0.38= -0.38 deg.c

f) Boiling point elevation = 2*0.56*0.102564=0.1148

Boiling point =100+0.1148= 100.1148 deg.c

g) i=1 for glycerol, boiling point = 78.4 ( boiling point of ethanol)+1*1.22*0.2 =78.644 deg.c

h) freezing point = -114.6 ( Freezing point for ethanol)- 0.2*1*1.99=-115 deg,c

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