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Question 1 of 9

Ringer's Solution for Injection contains 322 micrograms of Calcium Chloride Dihydrate in each millilitre. What is the millimolarity of the Calcium Chloride Dihydrate (CaCl2.2H2O)? Answer to 4 significant
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mM
The millimolarity of Calcium Chloride Dihydrate is 2.189 mM.
Millimolarity =     Number of Millimoles    
Volume of Total System (dm3 )
RMM CaCl2.2H2O = 147.1
Millimoles of CaCl2.2H2O = 1000 x Mass/RMM = 1000 x 0.000322 / 147.1 millimoles
= 0.0021889 millimoles

1 ml = 0.001 dm3
Millimolarity CaCl2.2H2O = 0.0021889 mmoles/dm3
0.001
Millimolarity CaCl2.2H2O = 2.1889 = 2.189 mmoles/dm3 to 4 sig figs

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Question 2 of 9

You need to prepare 300 cm3 of 0.7500 M Potassium Permanganate (KMnO4). What mass in grams of Potassium Permanganate would you weigh out?

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grams
You would weigh out 35.55 grams of Potassium Permanganate.

Number of Moles = Molarity x Volume of Total System (in dm3 )
300 cm3 = 0.3 dm3
Moles of Potassium Permanganate = 0.7500 x 0.3 = 0.225 moles

RMM Potassium Permanganate = 158.0

Mass of Potassium Permanganate = Moles x RMM
= 0.225 x 158.0 g
Mass of Potassium Permanganate = 35.55 g

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Question 3 of 9

During a titration of Sodium Hydroxide and Hydrochloric Acid, 26.7 cm3 of 0.9980 M Sodium Hydroxide were used. How many millimoles of Hydrochloric Acid were there? Answer to 4 significant figures.
NaOH + HCl ----» NaCl + H2O
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mmoles
26.65 millimoles of Hydrochloric Acid were used.
Find the number of moles of Sodium Hydroxide used.
Number of Moles = Molarity x Volume (dm3 )

Moles of NaOH used = 0.9980 x 0.0267 = 0.026646 moles
Millimoles of NaOH used = 1000 x 0.026646 = 26.646 millimoles

You can see from the equation that for every one mole of Sodium
Hydroxide used, one mole of Hydrochloric Acid is used.
Therefore:
Millimoles of HCl Used = Millimoles of NaOH Used
Millimoles of HCl Used = 26.646
Millimoles of HCl Used = 26.65 millimoles to 4 sig figs

Find out how many moles of Sodium Hydroxide were used. Then find out how many moles of Hydrochloric Acid are used for every one mole of Sodium Hydroxide used

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Question 4 of 9

What volume in ml of 0.9820 M Potassium Hydroxide is needed to just react with 16.2 ml of 0.4970 M Sulphuric Acid? Answer to 3 significant figures.

2 KOH + H2SO4 -----» K2SO4 + 2H2O
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ml
16.4 ml of Potassium Hydroxide are needed.
Find the number of moles of Sulphuric Acid used.
Number of Moles = Molarity x Volume (dm3 )
Moles of Sulphuric Acid used = 0.4970 x 0.0162 = 0.0080514 moles
You can see from the equation that for every one mole of Sulphuric Acid used, two moles of Potassium Hydroxide are needed.
Therefore:
Moles of KOH Needed    = 2 x Moles of Sulphuric Acid Used
Moles of KOH Needed    = 2 x 0.0080514 = 0.0161028 moles
Volume of KOH needed = Moles/Molarity = 0.0161028 x 0.9820 = 0.01639 dm3
Volume of KOH needed = 0.01639 x 1000 ml = 16.39 ml = 16.4 ml to 3 sig figs.

You need to find how many moles of Sulphuric Acid were used and then find how many moles of Potassium Hydroxide are needed.

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Question 5 of 9

50 ml of 0.8000 M Succinic Acid is mixed with 250 ml of 0.4000 Sodium Hydroxide. How many moles of Disodium Succinate are produced? Answer to 2 significant figures.

(CH2COOH)2 + 2 NaOH -----»(CH2COONa)2 + 2 H2O
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moles
0.040 moles of Disodium Succinate are produced.
Find out which of the two reactants is limiting the reaction - find out how many moles of each reactant are present.
Number of Moles = Molarity x Volume (in dm3 )
Moles of Succinic Acid = 0.8000 x 0.05 = 0.04 moles
Moles of Sodium Hydroxide = 0.4000 x 0.25 = 0.1 moles
From the equation you can see that for every 1 mole of Succinic Acid used, two moles of Sodium Hydroxide are used.
If Succinic Acid was limiting, the moles of Sodium Hydroxide needed would be 2 x 0.04 ie 0.08 moles. There are 0.1 moles of Sodium Hydroxide ie more than enough: thus Succinic Acid is the limiting reactant.
From the equation you can see that for every one mole of Succinic Acid used, one mole of Disodium Succinate is produced. Therefore:
Moles of Disodium Succinate produced = 0.04 moles = 0.040 to 2 sig figs.

Find out or Succinic Acid and Sodium Hydroxide limits the reaction. Then find out how many moles of Disodium Succinate would be produced.

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Question 6 of 9

350 ml of Disodium Succinate solution is made using 22.715 grams of Succinic Acid (RMM 118.0) and 15.4 grams of Sodium Hydroxide (RMM 40.0). What is the molarity of Disodium Succinate in the solution? Answer to 4 significant figures.
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M
The solution is 0.5500 M Disodium Succinate.
(CH2COOH)2 + 2 NaOH -----» (CH2COONa)2 + 2 H2O
Find out whether either of Disodium Succinate or Sodium Hydroxide limits the reaction by finding how many moles of each there are.
Moles of Succinic Acid         = 22.715/118.0 = 0.1925 moles
Moles of Sodium Hydroxide = 15.4/40.0       = 0.385 moles
1 mole of Succinic Acid reacts with 2 moles of Sodium Hydroxide. If Succinic Acid limited the reaction, the minimum amount of Sodium Hydroxide required would be 2 x 0.1925 moles ie 0.385 moles. All of the Succinic Acid and Sodium Hydroxide reacts.
1 mole of Succinic Acid produces 1 mole of Disodium Succinate. Thus 0.1925 moles of Succinic Acid produces 0.1925 moles of Disodium Succinate.
Molarity of Disodium Succinate Solution = Moles/Volume
= 0.1925/0.35 = 0.55 = 0.5500 to 4 sig figs

(CH2COOH)2 + 2 NaOH -----» (CH2COONa)2 + 2 H2O
Find out whether Succinic Acid or Sodium Hydroxide limits the reaction by finding how many moles of each there are present.

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Question 7 of 9

Up to what volume in ml would you need to make a solution if the solution was to contain 100 mg of Sodium Hydroxide (RMM 40.0) and have a final concentration of 100 mM Sodium Hydroxide?
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ml
The solution would be made up to 25 ml.
100 mg = 0.1 g

Moles of NaOH        = Mass/RMM = 0.1/40.0 = 0.0025 moles
Millimoles of NaOH = 1000 x 0.0025 = 2.5 millimoles

Volume = Number of Millimoles
Millimolarity

Volume of Solution = 2.5
100

Volume of Solution = 0.025 dm3 = 1000 x 0.025 ml = 25 ml

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Question 8 of 9

A solution contains 300 mg of Glacial Acetic Acid (Ethanoic Acid, RMM 60.05) in 100 ml of solution. What is the molarity of the solution? Answer to 4 significant figures.
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M
The solution is 0.04996 M Acetic Acid.
300 mg = 0.3 g
Moles of Acetic Acid = Mass / RMM
Moles of Acetic Acid = 0.3 / 60.05 = 0.0049958 moles

Molarity = Number of Moles
Volume (dm3 )

Molarity of Acetic Acid Solution = 0.0049958 M
0.1
Molarity of Acetic Acid Solution = 0.049958 M = 0.04996 M to 4 sig figs
NOTE: neither of the zeros are significant.

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Question 9 of 9

What volume in cm3 of 0.09890 M Hydrochloric Acid is needed to produce 953 mg of Magnesium Chloride in the reaction below if Magnesium Hydroxide is in excess? Answer to 3 significant figures.
Mg(OH)2 + 2 HCl -----» MgCl2 + 2 H2O
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cm3
202 cm3 Hydrochloric Acid are needed.
  1. 202 cm3 Hydrochloric Acid are needed.
    953 mg = 0.953 g            RMM Magnesium Chloride = 95.3
    Moles of Magnesium Chloride = Mass/RMM = 0.953/95.3 = 0.01 moles
  2. Find how many moles of Hydrochloric Acid are needed.
    You can see from the equation that for every one mole of Magnesium Chloride produced, two moles of Hydrochloric Acid are used.
    Therefore to produce 0.01 moles Magnesium Chloride, 2 x 0.01 moles ie 0.02 moles of Hydrochloric Acid are needed.
    Therefore volume of HCl needed is:
    Volume of HCl needed = Moles / Molarity
    Volume of HCl needed = 0.02 / 0.09890 dm3 = 0.2022 dm3
    Volume of HCl needed = 1000 x 0.2022 cm3 = 202.2 cm3 = 202 cm3 to 3 sig figs.
  1. Find how many moles of Magnesium Chloride are needed.
  2. Find how many moles of Hydrochloric Acid are needed.

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