17 Marks: 1 What is the freezing point of water made by dissolving 10.74 g of ma
ID: 1087437 • Letter: 1
Question
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 °CIm Answer: 18 Marks: 1 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. Answer 19 Marks: 1 What is the freezing point of water made by dissolving 46.18 g of sugar (C12H22011) in 90.93 g of water? The freezing-point depression constant of water is 1.86 °C/m. Answer:Explanation / Answer
17)
Lets calculate molality first
Molar mass of MgCl2,
MM = 1*MM(Mg) + 2*MM(Cl)
= 1*24.31 + 2*35.45
= 95.21 g/mol
mass(MgCl2)= 10.74 g
use:
number of mol of MgCl2,
n = mass of MgCl2/molar mass of MgCl2
=(10.74 g)/(95.21 g/mol)
= 0.1128 mol
m(solvent)= 86.59 g
= 8.659*10^-2 kg
use:
Molality,
m = number of mol / mass of solvent in Kg
=(0.1128 mol)/(0.08659 Kg)
= 1.303 molal
i for MgCl2 is 3
lets now calculate Tf
Tf = i*Kf*m
= 3*1.86*1.3027
= 7.2692 oC
This is decrease in freezing point
freezing point of pure liquid = 0.0 oC
So, new freezing point = 0 - 7.2692
= -7.2692 oC
Answer: -7.269 oC
18)
Lets calculate molality first
Molar mass of NaCl,
MM = 1*MM(Na) + 1*MM(Cl)
= 1*22.99 + 1*35.45
= 58.44 g/mol
mass(NaCl)= 16.31 g
use:
number of mol of NaCl,
n = mass of NaCl/molar mass of NaCl
=(16.31 g)/(58.44 g/mol)
= 0.2791 mol
m(solvent)= 93.18 g
= 9.318*10^-2 kg
use:
Molality,
m = number of mol / mass of solvent in Kg
=(0.2791 mol)/(0.09318 Kg)
= 2.995 molal
i for NaCl is 2
lets now calculate Tf
Tf = i*Kf*m
= 2.0*1.86*2.9952
= 11.142 oC
This is decrease in freezing point
freezing point of pure liquid = 0.0 oC
So, new freezing point = 0 - 11.142
= -11.142 oC
Answer: -11.14 oC
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