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Question 1 of 5

What is the concentration expressed as % w/w of a 2.598 M solution of Sucrose (RMM 342 g) if the density of the solution is 1.332 g/ml? Answer to 1 decimal place.

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% w/w
The concentration of the Sucrose is 66.7 % w/w.
  1. Find the mass of Sucrose in 1000 ml of solution.
    1000 ml of solution contains 2.598 moles of Sucrose. Thus:
    Mass of Sucrose = Moles x RMM = 2.598 x 342 g = 888.516 grams
  2. Find the mass of 1000 ml of solution - use the given density of 1.332 g/ml.
    1 ml of solution weighs 1.332 g. Thus:
    1000 ml of solution weighs 1332 grams.
  3. Find the mass of Sucrose in 100 g of solution.
    1000 ml of solution contains 888.516 g of Sucrose. From 2 you know that:
    1332 g of solution contains 888.516 g of Sucrose. Thus:
    100 g of solution contains 100/1332 x 888.516 g = 66.70 g of Sucrose
The concentration of Sucrose is 66.7 g/100 g ie 66.7 % w/w to 1 dec pl.

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Question 2 of 5

What is the molality of a 0.2228 M aqueous solution of Chlorhexidine Gluconate (RMM 897.7) if the solution's density is 1.065 g/ml? Answer to 4 significant figures.

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molal
The molality of the solution is 0.2576 molal.
  1. Find the mass of Chlorhexidine Gluconate (CG) in 1000 ml of solution.
    1000 ml of solution contains 0.2228 moles of Chlorhexidine Gluconate. Thus:
    Mass of Chlorhexidine Gluconate = Moles x RMM = 0.2228 x 897.7 = 200.007 grams
  2. Find the mass of 1000 ml of solution - use the given density of 1.065 g/ml
    1 ml of solution has a mass of 1.065 g. Thus:
    1000 ml of solution has a mass of 1065 grams
  3. Find the mass of solvent in 1000 ml (1065 grams) of solution.
    Mass of solvent = Mass of Solution - Mass of Chlorhexidine Gluconate
    Mass of solvent = 1065 - 200.007 g = 864.993 grams
  4. Find the number of moles associated with 1000 g of solvent.
    864.992 g of solvent are associated with 0.2228 moles of CG. Thus:
    1000 g of solvent are associated with 1000/864.992 x 0.2228 moles = 0.25757 moles
The concentration of the solution is 0.2576 moles/1000 g ie 0.2576 molal to 4 sig figs.

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Question 3 of 5


What is the mole fraction of Formaldehyde (RMM 30.0) in a 12.96 molar aqueous solution of Formaldehyde if the density of the solution is 1.08 g/ml? (RMM of Water = 18.02). Answer to 4 decimal places.

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The mole fraction of Formaldehyde is 0.2525.
  1. Find the mass of Formaldehyde in 1000 ml of solution.
    1000 ml of solution contains 12.96 moles of Formaldehyde. Thus:
    Mass of Formaldehyde = Moles x RMM = 12.96 x 30.0 g = 388.8 grams
  2. Find the mass of 1000 ml of solution - use the given density of 1.08 g/ml.
    1 ml of solution has a mass of 1.08 g. Thus:
    1000 ml of solution has a mass of 1080 grams
  3. Find the mass of Water present in 1000 ml (1080 g) of solution.
    Mass of Water = Mass of Solution - Mass of Formaldehyde
    Mass of Water = 1080 - 388.8 g = 691.2 grams
  4. Find the mole fraction of Formaldehyde.
    Moles of Formaldehyde = 12.96 moles
    Moles of Water = Mass/RMM = 691.2/18.02 = 38.357
Mole Fraction of Formaldehyde = 12.96 /(12.96 + 38.357) = 0.25254 = 0.2525 to 4 dec pl

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Question 4 of 5

What is the normality of a 1.650 molar solution of Potassium Sulphate? Answer to 4 significant figures.

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N
The normality of the solution is 3.300 N.
  1. Find the mass of one equivalent of Potassium Sulphate.
    Mass of Equivalent = RMM/Valency = 174.3/2 = 87.15 grams
  2. Find the mass of Potassium Sulphate in 1000 ml of solution.
    1000 ml of solution contains 1.650 moles of Potassium Sulphate. Thus:
    Mass of Potassium Sulphate = Moles x RMM = 1.650 x 174.3 g = 287.595 grams
  3. Find the number of equivalents of Potassium Sulphate in 1000 ml of solution.
    Number of Equivalents = Mass of Potassium Sulphate / Mass of One Equivalent
    Number of Equivalents = 287.595/87.15 Eq = 3.3000 Eq
The concentration of the solution is 3.300 Eq/1000 ml ie 3.300 N to 4 sig figs.

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Question 5 of 5

What is the concentration expressed as % w/v of a 0.7500 molar solution of Potassium Permanganate (KMnO4)?
Answer to 2 decimal places.

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% w/v
The concentration of the solution is 11.85 % w/v.
  1. Find the mass of Potassium Permanganate present in 1000 ml of solution.
    1000 ml of solution contains 0.7500 moles of Potassium Permanganate. Thus:
    Mass of Potassium Permanganate = Moles x RMM = 0.7500 x 158 g = 118.50 grams
  2. Find the mass of Potassium Permanganate in 100 ml of solution.
    1000 ml of solution contains 118.50 g of Potassium Permanganate. Thus:
    100 ml of solution contains 100/1000 x 118.50 g of Potassium Permanganate
    100 ml of solution contains 11.850 g of Potassium Permanganate.
The concentration of Potassium Permanganate is 11.85 g/100 ml
ie 11.85 % w/v to 2 dec pl.

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The % w/v of any component is the number of grams of that component contained in 100 cm3 (or 100 ml) of the total product.
The molarity of any component is the number of moles
of that component in 1000 cm3 of the total system.
The molality of a component in a solution is the number of moles of that component associated with one kilogram of the SOLVENT.
The mole fraction of a component within a system is the ratio
of the number of moles of that component to the total number
of moles of all components within the system.
The Normality of a component is the number of gram equivalents of that componet in 1000 cm3 (or 1 litre) of the total system.
The % w/w of any component is the number of grams of that component contained in 100 grams of the total product.
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