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Question 1 of 5

What is the molarity of a 5.857 molal solution of Sucrose (RMM 342.0) if the density of the solution is 1.332 g/ml? Answer to 4 significant figures.

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M
The molarity of the solution is 2.598 M.
  1. Find the mass of Sucrose associated with 1000 g of solvent.
    5.857 moles of Sucrose are associated with 1000 g of solvent. Thus:
    Mass of Sucrose = Moles x RMM = 5.857 x 342.0 g = 2003.1 grams
  2. Find the volume of solution containing 2003.0 g of Sucrose.
    Total MASS of solution = Mass of Sucrose + Mass of Solvent = 2003.1 + 1000 = 3003.1 g
    Using the given density of 1.332 g/ml:
    1.332 g of solution has a volume of 1 ml. Thus:
    3003.0 g of solution has a volume of 3003.1/1.332 ml = 2254.58 ml.
  3. Find the number of moles of Sucrose in 1000 ml of solution.
    2254.58 ml of solution contains 5.857 moles of Sucrose. Thus:
    1000 ml of solution contains 1000/2254.58 x 5.857 moles = 2.5978 moles of Sucrose
The concentration of Sucrose is 2.598 moles/1000 ml ie 2.598 M to 4 sig figs.

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Question 2 of 5

What is the concentration expressed as % w/v of a 18.75 molal solution of Formaldehyde (RMM 30.0) if the density of the solution is 1.08 g/ml? Answer to 2 decimal places.

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% w/v
The concentration of the solution is 38.88 % w/v.
  1. Find the mass of Formaldehyde associated with 1000 g of solvent.
    18.75 moles of Formaldehyde are associated with 1000 g of solvent. Thus:
    Mass of Formaldehyde = Moles x RMM = 18.75 x 30 g = 562.5 g
  2. Find the VOLUME of solution containing 562.5 g of Formaldehyde.
    MASS of Solution = Mass of Solvent + Mass of Formaldehyde = 1000 g + 562.5 g
    MASS of Solution = 1562.5 g
    Use the given density of 1.08 g/ml to find the volume of 1562.5 g of solution.
    1.08 g of solution has a volume of 1 ml. Thus:
    1562.5 g of solution has a volume of 1562.5/1.08 g = 1446.75 ml
  3. Find what mass of Formaldehyde is in 100 ml of solution.
    1446.75 ml of solution contains 562.5 g of Formaldehyde. Thus:
    100 ml of solution contains 100/1446.75 x 562.5 g = 38.880 g of Formaldehyde.
The concentration of Formaldehyde is 38.88 g/100 ml ie 38.88 % w/v to 2 dec pl.

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Question 3 of 5


What is the concentration expressed as % w/w of a 1.340 molal solution of Sodium Chloride? Answer to 2 decimal places.

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% w/w
The concentration of Sodium Chloride is 7.27 % w/w.
  1. Find the mass of Sodium Chloride associated with 1000 g of solvent.
    1.34 moles of Sodium Chloride are associated with 1000 g of solvent. Thus:
    Mass of Sodium Chloride = Moles x RMM = 1.34 x 58.5 g = 78.39 g
  2. Find the total mass of solution.
    Mass of Solution = Mass of Sodium Chloride + Mass of Solvent
    Mass of Solution = 78.39 g + 1000 g = 1078.39 g
  3. Find the mass of Sodium Chloride in 100 g of solution.
    1078.39 g of solution contains 78.39 g of Sodium Chloride. Thus:
    100 g of solution contains 100/1078.39 g = 7.269 g of Sodium Chloride
The concentration of Sodium Chloride is 7.27 g/100 g ie 7.27 % w/w to 2 dec pl.

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Question 4 of 5

What is the mole fraction of Potassium Nitrate in a 0.4250 molal aqueous solution of Potassium Nitrate? (RMM of Water = 18.02). Answer to 4 decimal places.

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The mole fraction of Potassium Nitrate is 0.0076.
  1. Find the number of moles of water associated with 0.4250 moles of Potassium Nitrate.
    1000 g of water are associated with 0.4250 moles of Potassium Nitrate. Thus:
    Moles of Water = Mass / RMM = 1000 / 18.02 moles = 55.4938 moles
  2. Find the mole fraction of Potassium Nitrate.
    Mole Fraction of Potassium Nitrate =                 Moles of Potassium Nitrate                
    Moles of Potassium Nitrate + Moles of Water
    Mole Fraction of Potassium Nitrate =           0.4250          
    0.4250 + 55.4938
    Mole Fraction of Potassium Nitrate = 0.00760 = 0.0076 to 4 dec pl

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Question 5 of 5

What is the normality of a 0.2119 molal solution of Sodium Acid Phosphate BP (NaH2PO4.2H2O, RMM 156.0) if the density of the solution is 1.008 g/ml?
Answer to 4 significant figures.

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N
The normality of the solution is 0.2068 N.
  1. Find the mass of one equivalent of Sodium Acid Phosphate BP.
    Mass of Equivalent = RMM/Valency = 156.0/1 g = 156.0 g
  2. Find the mass of Sodium Acid Phosphate (SAP) associated with 1000 g of water.
    0.2119 moles of SAP are associated with 1000 g of water.
    Mass of SAP = Moles x RMM = 0.2119 x 156.0 g = 33.0564 g
  3. Find the volume of solution containing 33.0564 g of SAP.
    Mass of Solution = Mass of Water + Mass of SAP = 1000 g + 33.0564 g = 1033.0564 g
    Use the given density of 1.008 g/ml to find the volume of 1033.0564 g of solution.
    1.008 g of solution has a volume of 1 ml. Thus:
    1033.0564 g of solution has a volume of 1033.0564/1.008 ml = 1034.8575 ml
  4. Find the number of equivalents in 1024.8575 ml of solution.
    Number of Equivalents = Mass of SAP/Mass of Equivalent = 33.0564/156 Eq = 0.2119 Eq
  5. Find the number of equivalents in 1000 ml of solution
    1024.8575 ml of solution contains 0.2119 Eq of SAP. Thus:
    1000 ml of solution contains 1000/1024.8575 x 0.2119 Eq = 0.20676 Eq = 0.2068 Eq

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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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