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Question 1 of 6

How many milligrams of Bicarbonate are there in 150 ml of a solution of Bicarbonate which has a milliequivalence of 28? Answer to 1 decimal place.

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mg
There are 256.2 milligrams of Bicarbonate.
  1. Find the number of milliequivalents in 150 ml of solution.
    A concentration of 28 mEq / litre means that:
    1000 ml of solution contains 28 mEq of Bicarbonate. Thus:
    150 ml of solution contains 150/1000 x 28 mEq = 4.2 milliequivalents.
  2. Find the weight of 1 milliequivalent of Bicarbonate.
    The gram equivalent of Bicarbonate is RMM/1 = 61.0/1 g = 61.0 grams
    One milliequivalent of Bicarbonate weighs 61.0 milligrams.
  3. Find the mass of Bicarbonate present.
    Mass = Number of Milliequivalents x Weight of One Milliequivalent.
    Mass of Bicarbonate = 4.2 x 61.0 mg = 256.20 mg = 256.2 mg to 1 dec pl

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Question 2 of 6

You need to prepare 500 ml of a 0.005 N solution of Calcium ions. How many milligrams - to the nearest milligram - of Calcium Chloride Hexahydrate do you need?

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mg
274 milligrams of Calcium Chloride Hexahydrate are needed.
  1. Find how many grams of Calcium are needed.
    A concentration of 0.005 N means that:
    1000 ml of the solution is to contain 0.005 Equivalents of Calcium. Thus:
    500 ml of the solution is to contain 500/1000 x 0.005 Eq = 0.0025 Eq.
    1 equivalent of Calcium weighs 40.1/2 g = 20.05 g
    Thus 0.0025 Eq weigh 0.0025 x 20.0 g = 0.050125 g
  2. Find how many moles of Calcium are needed.
    Moles of Calcium = Mass/RMM = 0.050125/40.01 = 0.00125 moles
  3. 1 mole of Calcium is provided by 1 mole of Calcium Chloride Hexahydrate. Thus :
    0.00125 moles of Calcium are provided by 0.00125 moles.
    Mass of Calcium Chloride Hexahydrate needed = Moles x RMM = 0.00125 x 219 g
    Mass of Calcium Chloride Hexahydrate needed = 0.27375 g = 273.75 mg
    Mass needed is 274 milligrams to the nearest milligram.

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Question 3 of 6

How many grams of Succinic Acid (CH2COOH)2 do you need to produce 300 ml of a 0.4700 N solution of Sodium Succinate if there is already just enough Sodium Hydroxide present? Answer to 3 decimal places.
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g
You need 8.319 grams of Succinic Acid.
  1. The reaction involved is: (CH2COOH)2 + NaOH ––––» (CH2COONa)2 + 2 H22O
  2. Find what mass of Sodium Succinate is needed.
    A concentration of 0.4700 N means that:
    1000 ml of the solution is to contain 0.4700 Eq of Sodium Succinate. Thus:
    300 ml of solution is to contain 300/1000 x 0.4700 Eq = 0.141 Eq of Succinate.
    The gram equivalent of Sodium Succinate is RMM/2 = 162/2 = 81 grams
    Thus mass of Succinate needed = 0.141 x 81 g = 11.421 g
  3. Find the moles of Succinate needed.
    Moles = Mass/RMM = 11.421/162 = 0.0705 moles
  4. From the equation you can see that:
    1 mole of Succinate is produced from 1 mole of Succinic Acid. Thus:
    0.0705 moles of Succinate are produced from 0.0705 moles of Succinic Acid.
    Thus Mass of Succinic Acid needed = Moles x RMM = 0.0705 x 118 g = 8.3190 g
    Mass of Succinic Acid needed is 8.319 g to 3 dec pl.

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Question 4 of 6

How many millilitres of a Sodium Chloride solution of milliequivalence 110 do you need if you require 213 milligrams of Chloride? Answer to 1 decimal place.

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ml
You need 54.5 ml of the solution.
  1. Find the number of moles of Chloride required.
    Moles = Mass / RMM = 0.213 / 35.5 = 0.006 moles
  2. Find the number of moles of Sodium Chloride in 1 litre of solution.
    A concentration of 110 milliequivalence means that:
    1000 ml of solution contains 110 mEq of Sodium Chloride.
    1 equivalent of Sodium Chloride weighs RMM/1 = 58.5/1 g = 58.5 grams. Thus:
    1 milliequivalent weighs 0.0585 grams.
    1000 ml of solution contains 110 x 0.0585 g = 6.435 g of Sodium Chloride
    1000 ml of solution contains (Mass/RMM = 6.435/58.5) = 0.11 moles of NaCl
  3. 1 mole of Chloride ions are provided by 1 mole of Sodium Chloride. Thus:
    0.006 moles of Chloride are provided by 0.006 moles of Sodium Chloride.
  4. There are 0.11 moles of Sodium Chloride in 1000 ml of solution. Thus:
    There are 0.006 moles of Sodium Chloride in 0.006/0.11 x 1000 ml = 54.54 ml
    Volume of solution required is 54.5 ml to 1 dec pl.

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Question 5 of 6

You need to prepare 1.5 litres of a 142 milliequivalence solution of Sodium. How many grams of Sodium Chloride do you need? Answer to 2 decimal places.
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grams
You need 12.46 grams of Sodium Chloride.
  1. Find the mass of SODIUM needed.
    A concentration of 142 milliequivalence means that:
    1 litre of solution is to contain 142 mEq of Sodium. Thus:
    1.5 litres of solution are to contain 1.5/1 x 142 mEq = 213 mEq = 0.213 Eq of Sodium.
    1 equivalent of Sodium weighs 23.0 g. Thus:
    0.213 Eq of Sodium weighs 0.213 x 23.0 g = 4.899 g
    4.899 grams of Sodium are needed.
  2. Find the number of moles of Sodium needed.
    Moles = Mass/RMM = 4.899 / 23.0 = 0.213 moles.
  3. Find the mass of Sodium Chloride needed.
    You know that 1 mole of Sodium is provided by 1 mole Sodium Chloride. Thus:
    0.213 moles of Sodium are provided by 0.213 moles of Sodium Chloride.
    Mass of Sodium Chloride needed = Moles x RMM = 0.213 x 58.5 g = 12.460 g
    Mass of Sodium Chloride needed = 12.46 g to 2 dec pl.

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Question 6 of 6

25 ml of a 0.8000 N solution of Sodium Hydroxide is added to just enough Sulphuric Acid to completely react and produce 55 ml of solution. What is the normality of the resulting solution? Answer to 4 decimal places.
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N
The Normality of the resulting solution is 0.3636 N.
  1. The reaction involved is: H2SO4 + 2 NaOH ––––» Na2SO4 + 2 H20
  2. Find the number of moles of Sodium Hydroxide used.
    A concentration of 0.8000 N means that 1000 ml of solution contains 0.8000 Eq of NaOH.
    Thus: 25 ml of solution contains 25/1000 x 0.8000 Eq = 0.02 Eq of NaOH.
    1 equivalent of NaOH weighs 40.0/1 = 40.0 g. Thus:
    Mass of NaOH used = Number of Eq x Weight of Eq = 0.02 x 40.0 g = 0.8 g
    Moles of NaOH used = Mass/RMM = 0.8/40.0 = 0.02 moles
  3. Find the number of equivalents of Sodium Sulphate produced.
    From the equation you can see that 1 mole of NaOH produces 1/2 mole of Sodium Sulphate.
    Thus, 0.02 moles of NaOH produce 0.01 mole of Sodium Sulphate.
    Mass of Sodium Sulphate produced = Moles x RMM = 0.01 x 142 g = 1.42 g
    1 equivalent of Sodium Sulphate weighs RMM/2 = 142/2 = 71 grams
    Equivalents of Sodium Sulphate produced = 1.42/71 = 0.02 Eq.
  4. 55 ml of solution contains 0.02 Eq. Thus:
    1000 ml of solution contains 1000/55 x 0.02 Eq = 0.36363 Eq = 0.3636 Eq to 4 dec pl.
    The concentration is 0.3636 Eq/litre which is the same as a Normality of 0.3636.

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