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Question 1 of 6
How many moles of Potassium Chloride (RMM 74.56) are there in 225 ml of solution if the concentration of Potassium Chloride is 1 in 8? Answer to 4 significant figures.
moles
- Find how many GRAMS of Potassium Chloride there are in 225 ml of solution.
The concentration of KCl is 1 part in 8 ie 1 g of KCl in 8 ml of solution.
Thus 8 ml of solution contains 1 g of Potassium Chloride
Thus 225 ml of solution contains 225/8 x 1 g = 28.125 g of Potassium Chloride - Find how many MOLES of Potassium Chloride are present in 225 ml of solution.
Moles of Potassium Chloride = Mass / RMM
Moles of Potassium Chloride = 28.125 . 74.56 moles = 0.37721 moles
Moles of Potassium Chloride = 0.3722 moles to 4 sig figs
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Question 2 of 6
An aqueous solution of the liquid Benzyl Alcohol contains 1 in 16 Benzyl Alcohol. How many millilitres of Benzyl Alcohol are there in 800 ml of the solution?
ie there is 1 ml of Benzyl Alcohol in 16 ml of solution.
Thus 16 ml of solution contains 1 ml of Benzyl Alcohol
Thus 800 ml of solution contains 800 x 1 ml of Benzyl Alcohol
16
800 ml of solution contains 50 ml of Benzyl Alcohol.
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Question 3 of 6
A solution contains Carbon Dioxide in water, 1 part in 6000. How many millimoles of Carbon Dioxide are there in 3000 grams of the solution? Answer to 4 significant figures.
- Find the MASS of Carbon Dioxide in 3000 g of solution.
The concentration of CO2 is 1 part in 6000 ie 1 g of CO2 in 6000 g of solution.
Thus 6000 g of solution contains 1 g of CO2
Thus 3000 g of solution contains 3000 / 6000 x 1 g = 0.5 g of CO2
- Find the MOLES of Carbon Dioxide present.
RMM of Carbon Dioxide = 44.0
Moles of Carbon Dioxide = Mass / RMM
Moles of Carbon Dioxide = 0.5 / 44.0 moles = 0.011363 moles
Millimoles of Carbon Dioxide = 1000 x 0.011363 = 11.363 mmoles
Millimoles of Carbon Dioxide = 11.36 mmoles to 4 sig figs
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Question 4 of 6
An aqueous solution of Sucrose (RMM 342.0) contains 1 in 1.1256 Sucrose. If the density of the solution is 1.332 g/ml, how many moles of Sucrose are there in 500 grams of solution? Answer to 4 significant figures.
- Find the volume of 500 grams of solution. Use the given density of 1.332 g/ml.
1.332 g of solution has a volume of 1 ml.
Thus, 500 g of solution has a volume of 500/1.332 x 1 ml = 375.375 ml - Find the MASS of Sucrose in 375.375 ml of solution.
The concentration is 1 in 1.1256 Sucrose ie 1 g of Sucrose in 1.1256 ml of solution.
Thus 1.1256 ml of solution contains 1 g of Sucrose.
Thus 375.375 ml of solution contains 375.375/1.1256 x 1 ml = 333.489 ml of Sucrose
- Find the number of MOLES of Sucrose.
Moles of Sucrose = Mass / RMM
Moles of Sucrose = 333.489 / 342.0 = 0.97511 moles
Moles of Sucrose = 0.9751 moles to 4 sig figs
- Find the volume of 500 grams of solution.
Use the given density of 1.332 g/ml.
1.332 g of solution has a volume of 1 ml. - Find the mass of Sucrose in the above volume of solution.
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Question 5 of 6
A pharmacist needs to prepare 400 ml of a 1 in 20 Sodium Chloride solution. The pharmacist has a 1 in 5 Sodium Chloride solution available. How many millilitres of the 1 in 5 solution does the pharmacist need to prepare the 400 ml of 1 in 20 solution?
- Find the MASS of Sodium Chloride needed for the 1 in 20 solution.
A concentration of 1 in 20 NaCl means that:
20 ml of solution contains 1 g of Sodium Chloride
Thus 400 ml of solution contains 400/20 x 1 g = 20 g of Sodium Chloride
Thus 20 g of Sodium Chloride are needed.
- Find the VOLUME of 1 in 5 NaCl solution containing 20 g of Sodium Chloride.
A concentration of 1 in 5 NaCl means that there are:
1 g of Sodium Chloride in 5 ml of solution
Thus 20 g of Sodium Chloride are in 20/1 x 5 ml
Thus 20 g of Sodium Chloride are in 100 ml of solution.
- Find the mass of Sodium Chloride needed for the 1 in 20 solution.
- Find the volume of the 1 in 5 solution containing the above mass of Sodium Chloride.
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Question 6 of 6
A pharmacist needs to prepare 250 ml of a Sodium Salicylate solution containing 1 part in 80 Sodium Salicylate. The pharmacist has a 1 in 10 Sodium Salicylate solution. How many millilitres of this solution does the pharmacist need to prepare the 250 ml of 1 in 80 Sodium Salicylate solution? Answer to 2 decimal places.
- Find the MASS of Sodium Salicylate needed for the 1 in 80 solution.
A concentration of 1 in 80 Sodium Salicylate means that:
80 ml of solution contains 1 g of Sodium Salicylate
Thus 250 ml of solution contains 250/80 x 1 g = 3.125 g of Sodium Salicylate
- Find the VOLUME of 1 in 10 solution containing 3.125 g of Sodium Salicylate.
A concentration of 1 in 10 Sodium Salicylate means that there is:
1 g of Sodium Salicylate in 10 ml of solution
Thus 3.125 g of Sodium Salicylate are in 3.125 x 10 ml
Thus 3.125 g of Sodium Salicylate are in 31.25 ml of solution to 2 dec pl
- Find the mass of Sodium Salicylate needed for the 1 in 80 solution.
- Find the volume of the 1 in 10 Sodium Salicylate solution containing the above mass of Sodium Salicylate.
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