15 Marks: 1 What is the boiling point of water made by dissolving 44.44 g of sug

Question

15 Marks: 1 What is the boiling point of water made by dissolving 44.44 g of sugar (C12H22011) in 85.94 g of water? The boiling-point elevation constant of water is 0.52 °Clm Answer: 16 What is the freezing point of water made by dissolving 24.06 g of ethylene glycol (CH2(OH)CH2(OH)) in 87.68 g of water? The freezing-point depression constant of water is 1.86 °Clm. Marks: 1 Answer: 17 Marks: 1 What is the freezing point of water made by dissolving 10.74 g of magnesium chloride in 86.59 g of water? The freezing-point depression constant of water is 1.86 °Clm. Answer: 18 What is the freezing point of water made by dissolving 16.31 g of sodium chloride in 93.18 g of water? The freezing-point depression constant of water is 1.86 °C/m. Marks: 1 Answer: 19 What is the freezing point of water made by dissolving 46.18 g of sugar (C12H2201) in 90.93 g of water? The freezing-point depression constant of water is 1.86 °C/m. Marks: 1 Answer:

Explanation / Answer

15)
Lets calculate molality first

Molar mass of C12H22O11,
MM = 12*MM(C) + 22*MM(H) + 11*MM(O)
= 12*12.01 + 22*1.008 + 11*16.0
= 342.296 g/mol


mass(C12H22O11)= 44.44 g

use:
number of mol of C12H22O11,
n = mass of C12H22O11/molar mass of C12H22O11
=(44.44 g)/(342.296 g/mol)
= 0.1298 mol

m(solvent)= 85.94 g
= 8.594*10^-2 kg

use:
Molality,
m = number of mol / mass of solvent in Kg
=(0.1298 mol)/(0.08594 Kg)
= 1.511 molal

lets now calculate Tb
Tb = Kb*m
= 0.52*1.5107
= 0.7856 oC

This is increase in boiling point
boiling point of pure liquid = 100.0 oC
So, new boiling point = 100 + 0.7856
= 100.79 oC
Answer: 100.79 oC

16)
Lets calculate molality first

Molar mass of C2H6O2,
MM = 2*MM(C) + 6*MM(H) + 2*MM(O)
= 2*12.01 + 6*1.008 + 2*16.0
= 62.068 g/mol


mass(C2H6O2)= 24.06 g

use:
number of mol of C2H6O2,
n = mass of C2H6O2/molar mass of C2H6O2
=(24.06 g)/(62.068 g/mol)
= 0.3876 mol

m(solvent)= 87.68 g
= 8.768*10^-2 kg

use:
Molality,
m = number of mol / mass of solvent in Kg
=(0.3876 mol)/(0.08768000000000001 Kg)
= 4.421 molal

lets now calculate Tf
Tf = Kf*m
= 1.86*4.4211
= 8.2232 oC

This is decrease in freezing point
freezing point of pure liquid = 0.0 oC
So, new freezing point = 0 - 8.2232
= -8.223 oC
Answer: -8.223 oC

Only 1 question at a time please

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.