Chapter13 UOC 1) An aqueous solution is made by dissolving 18.7 grams of aluminu
ID: 994126 • Letter: C
Question
Chapter13 UOC
1) An aqueous solution is made by dissolving 18.7 grams of aluminum chloride in 388 grams of water.
The molality of aluminum chloride in the solution is ____________ m.
2) In the laboratory you are asked to make a 0.738 m magnesium nitrate solution using 315 grams of water.
How many grams of magnesium nitrate should you add?
____________________ grams.
3)In the laboratory you are asked to make a 0.375 m cobalt(II) bromide solution using 17.7 grams of cobalt(II) bromide.
How much water should you add?
___________________grams
Thank you.
Explanation / Answer
Answer – 1) We are given mass of AlCl3 = 18.7 g , mass of water = 388 g
First we need to calculate the moles of aluminum chloride
Moles of AlCl3 = 18.7 g / 133.34 g.mol-1
= 0.140 moles
We know,
Molality = moles of solute / kg of solvent
Molality of AlCl3 = 0.140 moles / 0.388 kg
= 0.361 m
So, the molality of aluminum chloride in the solution is 0.361 m
2) We are given, molality of Mg(NO3)2 = 0.738 m, mass of water = 315 g
We know,
Molality = moles of solute / kg of solvent
Moles of solute = molality * kg of solvent
Moles of Mg(NO3)2 = 0.738 m * 0.315 kg
= 0.232 moles
Now mass of Mg(NO3)2
We know
Mass = moles * molar mass
Mass of Mg(NO3)2 = 0.232 moles * 148.3 g/mol
= 34.5 g
34.5 grams of magnesium nitrate should you add.
3) We are given, mass of cobalt (II) bromide = 17.7 g , molality of cobalt (II) bromide = 0.375 m
We need to calculate the moles of cobalt (II) bromide
Moles of CoBr2 = 17.7 g / 218.74 g.mol-1
= 0.0809 moles
We know,
Molality = moles of solute / kg of solvent
Mass of solvent, kg = moles of solute / molality
= 0.0809 moles / 0.375 m
= 0.216 kg
= 216 g
So, 216 g water should you add
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