Question

Assuming the van't hoff factor of FeCl3 is 3.4. Calculate the mass of the solute required to produce the aqueous solution of iron chloride that has 300g of water and a boiling point of 103 degrees celcius.

Answer #1

Boiling point elevation

T = i x Kb x m

(103 - 100) = 3°C = 3.4 x 0.512 °C/m x moles of solute /mass of water

moles of solute /mass of water = 1.723

moles of solute = 1.723 x 0.300 kg

= 0.517 moles

Mass of iron chloride = moles x molecular weight

= 0.517 moles x 162.206 g/mol

= 83.86 g

Use the van't Hoff factors in the table below to calculate each
colligative property. Table. Van't Hoff Factors at 0.05 m
Concentration in Aqueous Solution Solute i Expected i Measured
Nonelectrolyte 1 1 NaCl 2 1.9 MgSO4 2 1.3 MgCl2 3 2.7 K2SO4 3 2.6
FeCl3 4 3.4 part b the osmotic pressure of a 0.084 M potassium
sulfate solution at 298 K Part c the boiling point of a 1.12 % by
mass magnesium chloride solution

Use the van't Hoff factors in the table below to calculate each
colligative property.
Table. Van't Hoff Factors at 0.05 m Concentration in
Aqueous Solution
Solute
i Expected
i Measured
Nonelectrolyte
1
1
NaCl
2
1.9
MgSO4
2
1.3
MgCl2
3
2.7
K2SO4
3
2.6
FeCl3
4
3.4
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Express your answer using three significant figures
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Use the van't Hoff factors in the table below to calculate each
colligative property.
Table. Van't Hoff Factors at 0.05 m Concentration in Aqueous
Solution
Solute
i Expected
i Measured
Nonelectrolyte
1
1
{\rm NaCl}
2
1.9
{\rm MgSO_4}
2
1.3
{\rm MgCl_2}
3
2.7
{\rm K_2SO_4}
3
2.6
{\rm FeCl_3}
4
3.4
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