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Question 1 of 5
The mole fraction of Sucrose in Syrup BP is 0.0954. If the density of the Syrup is 1.332 g/ml, what is the molarity of Sucrose in the Syrup? Answer to 4 significant figures. (RMM of Water = 18.02).
- For every 1 mole of ALL components present:
Number of moles of Sucrose = 0.0954 moles
Number of moles of Water = 1 - 0.0954 = 0.9046 moles -
Find the mass of solution when there is a total of 1 mole of all components present.
Mass of Sucrose = Moles x RMM = 0.0954 x 342.0 g = 32.6268 g
Mass of Water = Moles x RMM = 0.9046 x 18.02g = 16.3008 g
Mass of Solution = Mass of Sucrose + Mass of Water = 32.6268 + 16.3008 g = 48.9276 g -
Find the volume of 48.0276 g of solution - use the given density of 1.332 g/ml.
1.332 g of solution has a volume of 1 ml. Thus:
48.9276 g of solution has a volume of 48.9276/1.332 ml = 36.7324 ml -
Find the number of moles of Sucrose in 1000 ml of solution.
36.7324 ml of solution contains 0.0954 moles of Sucrose. Thus:
1000 ml of solution contains 1000/36.7324 x 0.0954 moles = 2.5971 moles
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Question 2 of 5
The mole fraction of Potassium Chloride in an aqueous solution is 0.0025. What is the molality of the aqueous solution? (RMM of Water = 18.02) Answer to 4 significant figures.
- For every 1 mole of ALL components present:
Number of moles of Potassium Chloride = 0.0025 moles
Number of moles of Water = 1 - 0.0025 = 0.9975 moles -
Find the mass of 0.9975 moles of Water.
Mass of Water = Moles x RMM = 0.9975 x 18.02 g = 17.97495 g -
Find the moles of Potassium Chloride associated with 1000 g of Water.
17.97495 g of Water are associated with 0.0025 moles of Potassium Chloride. Thus:
1000 g of Water are associated with 1000/17.97495 x 0.0025 moles of KCl
1000 g of Water are associated with 0.13908 moles of KCl
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Question 3 of 5
The mole fraction of Chlorhexidine Gluconate (RMM 897.7) in an aqueous solution is 0.00462. The density of the solution is 1.065 g/ml. What is the concentration expressed as % w/v of Chlorhexidine Gluconate? (RMM of Water = 18.02). Answer to 1 decimal place.
- For every 1 mole of ALL components present:
Moles of Chlorhexidine Gluconate = 0.00462 moles
Moles of Water = 1 - 0.00462 = 0.99538 moles -
Find the total mass of solution containing 1 mole of all components.
Mass of Chlorhexidine = Moles x RMM = 0.00462 x 897.7 g = 4.1473 g
Mass of Water = Moles x RMM = 0.99538 x 18.02 g = 17.9367 g
Total mass of solution = 4.1473 g + 17.9367 g = 22.0840 g -
Find the volume of solution containing 1 mole of all components.
1.065 g of solution has a volume of 1 ml. Thus:
22.0840 g of solution has a volume of 22.0840/1.065 ml = 20.73615 ml -
Find the mass of Chlorhexidine Gluconate in 100 ml of solution.
20.73615 ml of solution contains 4.1473 g of Chlorhexidine Gluconate. Thus:
100 ml of solution contains 100/20.73615 x 4.1473 g = 20.00 g of Chlorhexidine.
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Question 4 of 5
The mole fraction of Sodium Chloride in an aqueous solution of Sodium Chloride is 0.0056. What is the concentration expressed as % w/w of Sodium Chloride? (RMM of Water = 18.02). Answer to 2 decimal places.
- For every 1 mole of ALL components present:
Moles of Sodium Chloride = 0.0056 moles
Moles of Water = 1 - 0.0056 = 0.9944 moles -
Find the mass of solution containing a total of 1 mole of components.
Mass of Sodium Chloride = Moles x RMM = 0.0056 x 58.5 g = 0.3276 g
Mass of Water = Moles x RMM = 0.9944 x 18.02 g = 17.919 g
Total Mass of Water = Mass of NaCl + Mass of Water = 0.3276 + 17.919 g = 18.2466 g -
Find the mass of Sodium Chloride in 100 g of solution.
18.2466 g of solution contains 0.3276 g of Sodium Chloride. Thus:
100 g of solution contains 100/18.2466 x 0.3276 g = 1.795 g of Sodium Chloride.
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Question 5 of 5
The mole fraction of Sodium Acid Phosphate BP (RMM 156.0) in an aqueous solution is 0.0038. What is the normality of the solution? The density of the solution is 1.008 g/ml. Answer to 4 significant figures.
- For every 1 mole of ALL components:
Moles of Sodium Acid Phosphate = 0.0038 moles
Moles of Water = 1 - 0.0038 = 0.9962 moles -
Find the mass of solution containing a total of 1 mole of components.
Mass of SAP = Moles x RMM = 0.0038 x 156.0 g = 0.5928 g
Mass of Water = Moles x RMM = 0.9962 x 18.02 g = 17.9515 g
Mass of Solution = Mass of SAP + Mass of Water = 0.5928 g + 17.9515 g = 18.5443 g -
Find the volume of 18.5443 g of solution - use the given density of 1.008 g/ml.
1.008 g of solution has a volume of 1 ml. Thus:
18.5443 g of solution has a volume of 18.5443/1.008 ml = 18.3971 ml -
Find the mass of Sodium Acid Phosphate in 1000 ml of solution.
18.3971 ml of solution contains 0.5928 g of SAP. Thus:
1000 ml of solution contains 1000/18.3971 x 0.5928 g = 32.222 g of SAP. -
Find the number of equivalents of SAP in 1000 ml of solution. One equivalent = 156.0 g
Number of equivalents = Mass of SAP/Mass of Equivalent = 32.333/156 = 0.20655 Eqs
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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.
