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Question 1 of 5
A solution of Sodium Chloride has a normality of 0.3650 N. What is the molarity of this solution? Answer to 4 significant figures.
- Find the mass of one equivalent of Sodium Chloride.
Mass of Equivalent = RMM/Valency = 58.5/1 g = 58.5 g -
Find the mass of Sodium Chloride in 1000 ml of solution.
1000 ml of solution contains 0.3650 equivalents of Sodium Chloride.
Mass of Sodium Chloride = Number of Equivalents x Mass of Equivalent
Mass of Sodium Chloride = 0.3650 x 58.5 g = 21.3525 g -
Find the number of moles of Sodium Chloride in 1000 ml of solution.
Number of Moles = Mass/RMM = 21.3525/58.5 = 0.36500 moles
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Question 2 of 5
The normality of a solution of Sodium Acid Phosphate BP is 0.2068 N. What is the molality of the solution? The density of the solution is 1.008 g/ml. Answer to 4 significant figures.
- Find the mass of Sodium Acid Phosphate (SAP) in 1000 ml of solution.
Mass of SAP Equivalent = RMM/Valency = 156.0/1 g = 156.0 g
Mass of SAP = Number of Equivalents x Mass of Equivalent = 0.2068 x 156 g
Mass of SAP = 32.2608 g -
Find the number of moles of Sodium Acid Phosphate present in 1000 ml of solution.
Moles of SAP = Mass/RMM = 32.2608/156.0 = 0.2068 moles -
Find the mass of solvent in 1000 ml of solution.
Mass of 1000 ml of solution = 1000 x 1.008 g = 1008 g
Mass of Solvent = Mass of Solution - Mass of SAP = 1008 - 32.2608 g = 975.7392 g -
Find the number of moles of SAP associated with 1000 g of solvent.
975.7392 g of solvent are associated with 0.2068 moles of SAP. Thus:
1000 g of solvent are associated with 1000/975.7392 x 0.2068 = 0.21194 moles of SAP.
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Question 3 of 5
The normality of a solution of Sodium Acid Phosphate BP (RMM 156.0) is 0.2068 N. What is the mole fraction of WATER? The density of the solution is 1.008 g/ml and the RMM of Water is 18.02. Answer to 4 decimal places.
- Find the number of moles of Sodium Acid Phosphate in 1000 ml of solution.
Mass of One Equivalent of SAP = RMM/Valency = 156.0/1 = 156.0 g
Mass of SAP = Number of SAP Eq x Mass of Eq = 0.2068 x 156.0 g = 32.2608 g
Moles of SAP = Mass/RMM = 32.2608/156.0 = 0.2068 moles -
Find the mass of 1000 ml of solution - use the given density of 1.008 g/ml
1 ml of solution weighs 1.008 g. THUS: 1000 ml of solution weighs 1008 g. -
Find the mass of water in 1008 g of solution.
Mass of Water = Mass of Solution - Mass of SAP = 1008 - 32.3608 g = 975.7392 g -
Find the mole fraction of water.
Moles of SAP = 0.2068 moles
Moles of Water = Mass/RMM = 975.7392/18.02 = 54.1476 moles
Mole Fraction of Water = 54.1476 = 0.99619 = 0.9962 to 4 sig figs
54.1476 + 0.2068
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Question 4 of 5
The normality of Sodium Acid Phosphate BP (RMM 156.0) in solution is 0.2068 N. The density of the solution is 1.008 g/ml. What is the concentration of Sodium Acid Phosphate expressed as % w/w? Answer to 2 decimal places.
- Find the mass of Sodium Acid Phosphate (SAP) in 1000 ml of solution.
1000 ml of solution contains 0.2068 equivalents of Sodium Acid Phosphate.
Mass of One Equivalent of SAP = RMM/Valency = 156.0/1 g = 156.0 g
Mass of SAP = Number of SAP Eqs x Mass of SAP Eq = 0.2068 x 156.0 g = 32.2608 g -
Find the mass of 1000 ml of solution - use the given density of 1.008 g/ml.
1 ml of solution has a mass of 1.008 g. Thus: 1000 ml has a mass of 1008 g. -
Find the mass of Sodium Acid Phosphate in 100 g of solution.
1008 g of solution contains 32.2608 g of Sodium Acid Phosphate. Thus:
100 g of solution contains 100/1008 x 32.2608 g = 3.200 g of Sodium Acid Phosphate.
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Question 5 of 5
The normality of a solution of Sodium Sulphate is 0.7340 N. What is the concentration - expressed as % w/v - of the solution? Answer to 2 decimal places.
- Find the mass of Sodium Sulphate in 1000 ml of solution.
1000 ml of solution contains 0.7340 equivalents of Sodium Sulphate.
Mass of One Equivalent of Sodium Sulphate = RMM/Valency = 142.1/2 g = 71.05 g
Mass of Sodium Sulphate = Number of Equivalents x Mass of One Equivalent
Mass of Sodium Sulphate = 0.7340 x 71.05 g = 52.1507 g -
Find the mass of Sodium Sulphate in 100 ml of solution.
1000 ml of solution contains 52.1507 g of Sodium Sulphate. Thus:
100 ml of solution contains 5.21507 g of Sodium Sulphate.
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of that component in 1000 cm3 of the total system.
of the number of moles of that component to the total number
of moles of all components within the system.
