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Question 1 of 7

19.5 grams of Sodium Chloride are dissolved in 100 moles of water (RMM 18.02). What is the molality of Sodium Chloride (RMM 58.5) in this solution. Answer to 4 significant figures.

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molal
The concentration of Sodium Chloride is 0.1850 molal.

  1. Find the mass of water involved:
    Mass of water = moles x RMM = 100 x 18.02 g = 1802 g = 1.802 kg

  2. Find the moles of NaCl involved:
    Moles of NaCl = Mass / RMM = 19.5 / 58.5 = 0.33333 moles
  3. Find the molality of NaCl
    Molality of NaCl = Moles of NaCl
    Mass of water
    Molality of NaCl = 0.33333 = 0.18497 moles/kg 0.1850 moles/ kg to 4 sig figs
    1.802

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Question 2 of 7

HYou need to prepare a 1.500 molal aqueous solution of Sodium Chloride. If you use 70 moles of water (RMM 18.02), how many grams of Sodium Chloride should you weigh out? Answer to 2 decimal places.

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grams
You should weigh out 110.69 grams of Sodium Chloride.

  1. Find the mass of water used:
    Mass of water = Moles x RMM = 70 x 18.02 g = 1261.4 g = 1.2614 kg
  2. Find the number of moles of Sodium Chloride needed.
    Moles of NaCl needed = Molality x Mass of Water
    Moles of NaCl needed = 1.500 x 1.2614 moles
    Moles of NaCl needed = 1.8921 moles
  3. Find the mass of NaCl needed:
    Mass of NaCl = Moles x RMM
    Mass of NaCl = 1.8921 x 58.5 = 110.687 g = 110.69 g to 2 dec pl

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Question 3 of 7

The concentration of Sodium Hydroxide in an aqueous solution is 0.4750 molal. How many MOLES of WATER (RMM 18.02) are associated with 224 milligrams of Sodium Hydroxide? Answer to 4 significant figures.

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moles
0.6542 moles of water are associated with the Sodium Hydroxide.
  1. Find the number of moles of Sodium Hydroxide involved:
    Convert mass in milligrams to mass in grams: 224 mg = 0.224 g
    Moles of NaOH = Mass / RMM = 0.224 / 40.0 = 0.0056 moles
  2. Find the mass of water associated with 0.0056 moles of NaOH:
    Mass of Water = Number of Moles of NaOH
    Molality of NaOH
    Mass of Water = 0.0056 = 0.0117894 kg = 11.7894 g
    0.4750
  3. Find the number of moles of water:
    Moles of Water = Mass / RMM
    Moles of Water = 11.7894 / 18.02 = 0.65424 moles
    Moles of Water = 0.6542 moles to 4 sig figs

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Question 4 of 7

An aqueous solution contains 17.7 grams of Succinic Acid ((CH2COOH)2) in 300 ml of solution. If the density of the solution is 0.820 g/ml, what is the molality of the solution? Answer to 4 significant figures.
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molal
The concentration of the Succinic Acid is 0.6570 molal.
  1. Find the mass of solution involved - use the given density of 0.820 g/ml
    1 ml of solution weighs 0.820 g thus:
    300 ml of solution weighs 300/1 x 0.820 g = 246 g
  2. Find the mass of water in the solution:
    Mass of water = Mass of Solution - Mass of Succinic Acid
    Mass of water = 246 - 17.7 g =228.3 g
  3. Find the number of moles of Succinic Acid present:
    Moles of Succinic Acid = Mass / RMM = 17.7 / 118.0 moles
    Moles of Succinic Acid = 0.15 moles
  4. Find the molality of Succinic Acid:
    Molality of Succinic Acid = Moles of Succinic Acid / Mass of Water in kgs
    Molality of Succinic Acid = 0.15 / 0.2283 =0.65703 moles/kg
    Molality of Succinic Acid = 0.6570 moles/kg to 4 sig figs
  1. Find the mass of the solution involved - use the given density of 0.820 g/ml thus:
    1 ml of solution weighs 0.820 g
  2. Find the mass of water in the solution.

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Question 5 of 7

You need to prepare 50 ml of a Strong Ammonia Solution containing Ammonia at a concentration of 23.447 molal, dissolved in water. The density of the solution should end up as 0.8965 g/ml. How many grams of Ammonia do you need? Answer to 2 decimal places.
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grams
You need 12.78 grams of Ammonia.
  1. Find the mass of Ammonia associated with 1000 grams of water.
    23.477 molal means that 23.477 moles of NH3 are associated with 1000 grams of water
    Mass of Ammonia associated with 1000 grams of water is:
    Moles x RMM = 23.447 x 17.0 g = 398.599 g
  2. Find the volume of solution containing 398.599 g of Ammonia. Use the given density.
    Mass of solution containing 398.599 g of Ammonia is:
    Mass of water + Mass of Ammonia = 1000 + 398.599 g = 1398.599 g
    Volume of 1398.599 g of SOLUTION is, using the density:
    0.8965 g of solution has a volume of 1 ml
    1398.599 g of solution has a volume of 1398.599 / 0.8965 x 1 ml = 1560.0658 ml
  3. Find the mass of Ammonia needed to prepare 50 ml of solution:
    1560.0658 ml solution contains 398.599 g of Ammonia
    Thus 50 ml contains 50 / 1560.0658 x 398.599 g ie 12.775 g of NH3 = 12.78 g to 2 dec pl
  1. Find the mass of Ammonia associated with 1000 grams of water.
  2. Find the volume of solution containing the above mass of Ammonia - given density of 0.8965 g/ml:
    0.8965 g of solution has a volume of 1 ml

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Question 6 of 7

How many moles of Glucose are associated with 10 moles of water (RMM 18.02) in a 0.8500 molal aqueous solution of Glucose? Answer to 4 significant figures.

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moles
0.1532 moles of Glucose are associated with 10 moles of water.
  1. Find the mass of water involved:
    Mass of water = Moles x RMM = 10 x 18.02 g = 180.2 g = 0.1802 kg
  2. Find the number of moles of Glucose:
    Moles of Glucose = Molality of Glucose x Mass of Water in kgs
    Moles of Glucose = 0.8500 x 0.1802 moles
    Moles of Glucose = 0.15317 moles
    Moles of Glucose = 0.1532 moles to 4 sig figs

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Question 7 of 7

What is the molality of Chlorhexidine Gluconate in aqueous Chlorhexidine Gluconate Solution if there are 30 grams of Chlorhexidine Gluconate (RMM 897.7) in 150 ml of the solution and the density of the solution is 1.065 g/ml? Answer to 4 significant figures.
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molal
The molality of Chlorhexidine Gluconate is 0.2576 molal.
  1. Find the mass of solution involved - use the given density of 1.065 g/ml:
    1 ml of solution weighs 1.065 g
    Thus 150 ml of solution weighs 150/1 x 1.065 g = 159.75 g
  2. Find the mass of water associated with 30 g Chlorhexidine Gluconate:
    Of the 159.75 g of solution, 30 g are Chlorhexidine Gluconate.
    Thus, 30 g of Chlorhexidine Gluconate are associated with 159.75 - 30 g of water
    30 g of Chlorhexidine Gluconate are associated with 129.75 g ie 0.12975 kg water
  3. Find the number of moles of Chlorhexidine Gluconate:
    Moles of Gluconate = Mass / RMM = 30 / 897.7 = 0.034187 moles
  4. Calculate the molality of Chlorhexidine Gluconate:
    Molality of Gluconate = Moles of Gluconate / Mass of Water in kgs
    Molality of Gluconate = 0.0334187 / 0.12975 = 0.25756 moles/kg
    Molality of Gluconate = 0.2576 moles/kg to 4 sig figs
  1. Find the mass of solution involved - use the given density of 1.065 g/ml
    1 ml of solution weighs 1.065 g
  2. Find the mass of water associated with 30 g of Chlorhexidine Gluconate

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