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Question 1 of 7

You have a saturated solution of Paracetamol in Ethanol (96%). The solubility of Paracetamol is 7 parts in Ethanol (96%). If the solution contains 420 ml of 96% Ethanol, how many grams of Paracetamol are there present?

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grams
There are 60 grams of Paracetamol.
The solubility indicates that 1 gram of Paracetamol
dissolves in 7 ml of 96 % Ethanol.
Since the solution is saturated:

7 ml of 96% Ethanol contains 1 g of Paracetamol

Thus 420 ml of 96% Ethanol contains  420  x 1 g of Paracetamol.
7

420 ml of 96 % Ethanol contains 60 grams of Paracetamol.

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Question 2 of 7

You have a saturated solution of Alum (RMM 474.4) in Glycerol. Alum is soluble in 3 parts of Glycerol. Assuming the solubility to be correct, how many millimoles of Alum are dissolved in 600 ml of Glycerol? Answer to 4 significant figures.

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mmoles
421.6 mmoles of Alum are dissolved.

  1. The solubility indicates that:
    1 g of Alum dissolves in 3 ml of Glycerol
    The solution is saturated and the solubility is assumed to be correct. Thus:
    3 ml of Glycerol contains 1 g of Alum
    Thus 600 ml of Glycerol contains 600 / 3 x 1 g ie 200 g of Alum.
  2. Find the number of moles of Alum present:
    Moles of Alum = Mass / RMM
    Moles of Alum = 200 / 474.4 moles = 0.42158 moles = 421.58 millimoles
    Millimoles of Alum = 421.6 mmoles to 4 sig figs

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Question 3 of 7

A saturated solution of Sucrose (RMM 342.0) contains 438.59 millimoles of Sucrose in 150 ml of water. What is the solubility of Sucrose in water? Answer to 1 decimal place.

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Sucrose is soluble in part(s) of water.
Sucrose is soluble in 1.0 parts of water.
  1. Find the mass of Sucrose dissolved in 150 ml of water.
    Convert millimoles to moles: 438.59 mmoles = 0.43859 moles
    Mass of Sucrose dissolved = Moles x RMM = 0.43859 x 342 g
    Mass of Sucrose dissolved = 149.99778 g
  2. Find in what volume of water 1 g of Sucrose is dissolved.
    149.99778 g of Sucrose are dissolved in 150 ml of water.
    Thus 1 g of Sucrose is dissolved in 1 / 149.99778 x 150 ml of water
    Thus 1 g of Sucrose is dissolved in 1.00 ml of water
    1 g of Sucrose is dissolved in 1.0 ml of water to 1 dec pl

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Question 4 of 7

Urea (CO(NH2)2) is soluble in 1 part of water. Assuming that this solubility is correct, how many moles of Urea are there in 225 ml of water if the solution is saturated? Answer to 4 significant figures.
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moles
There are 3.750 moles of Urea in 225 ml of water.
  1. The solubility means that:

    1 g of Urea dissolves in 1 ml of water
    Since the solution is saturated and the solubility is assumed to be correct:
    1 ml of water contains 1 g of Urea
    Thus 225 ml of water contains 225/1 x 1 g = 225 g of Urea

  2. Find the number of moles of Urea dissolved:

    RMM Urea = 60.0
    Moles of Urea = Mass / RMM = 225 / 60.0 = 3.75 moles
    Moles of Urea = 3.750 moles to 4 sig figs

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Question 5 of 7

A saturated solution of Ammonium Chloride (NH4Cl) contains 2.8037 moles of Ammonium Chloride in 405 ml of water. What is the solubility of Ammonium Chloride in water? Answer to 1 decimal place.
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Ammonium Chloride is soluble in parts of water.
Ammonium Chloride is soluble in 2.7 parts of water.
  1. Find the mass of Ammonium Chloride dissolved in 405 ml of water.

    Mass of Ammonium Chloride = Moles x RMM
    Mass of Ammonium Chloride = 2.8037 x 53.5 g = 149.99795 g

  2. Find in what volume of water 1 g of Ammonium Chloride dissolves:

    149.99795 g of Ammonium Chloride dissolves in 405 ml of water
    Thus 1 g of Ammonium Chloride dissolves in 1/149.99795 x 405 ml of water
    1 g of Ammonium Chloride dissolves in 2.70 ml of water
    1 g of Ammonium Chloride dissolves in 2.7 ml of water to 1 dec pl

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Question 6 of 7

A saturated solution of Dimercaprol (a liquid) contains 450 ml of the solvent Arachis Oil. Dimercaprol is soluble in 18 parts of Arachis Oil. Assuming the solution has this solubility, how many millilitres of Dimercaprol are present?

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ml
25 ml of Dimercaprol are present.
The solubility means that:
1 ml of Dimercaprol dissolves in 18 ml of Arachis Oil.

Since the solution is saturated and the solubility is assumed to be as above:

18 ml of Arachis Oil contains 1 ml of Dimercaprol
Thus 450 ml of Arachis Oil contains 450 / 18 x 1 ml of Dimercaprol
Thus 450 ml of Arachis Oil contains 25 ml of Dimercaprol.

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Question 7 of 7

The solid, Hydrocortisone (RMM 362.5) is soluble in 40 parts of 96 % Ethanol. If the density of 96 % Ethanol is 0.8038 g/ml, how many millimoles of Hydrocortisone are associated with 1000 grams of 96 % Ethanol in a saturated solution of Hydrocortisone with the above solubility? Answer to 4 significant figures.
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millimoles
85.80 mmoles of Hydrocortisone are associated with 1000 g of Ethanol.
  1. The solubility indicates that 1 g of Hydrocortisone dissolves in 40 ml of 96 % Ethanol.

    Since the solution is saturated and the solubility is assumed to be as above:
    40 ml of 96 % Ethanol are associated with 1 g of Hydrocortisone.

  2. Find out how much 40 ml of 96 % Ethanol weighs. Use the given density of 0.8038 g/ml

    1 ml of 96 % Ethanol weighs 0.8038 g
    Thus 40 ml of 96 % Ethanol weighs 40 x 0.8038 g = 32.152 g

  3. Find out what MASS of Hydrocortisone is associated with 1000 g of Ethanol

    32.152 g of Ethanol contains 1 g of Hydrocortisone
    Thus 1000 g of Ethanol contains 1000/32.152 x 1 g = 31.102264 g of Hydrocortisone

  4. Find out the number of moles of Hydrocortisone associated with 1000 g of Ethanol

    Number of Moles = Mass/RMM
    Number of Moles = 31.102264 / 362.5 = 0.0857993 moles = 85.799 millimoles
    Number of Moles = 85.80 millimoles to 4 sig figs

  1. Find out how much 40 ml of 96 % Ethanol weighs. Use the given density of
    0.8307 g/ ml:
    1 ml of 96% Ethanol weighs 0.8307 g
  2. Find out what mass of Hydrocortisone is associated with 1000 g of Ethanol.

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