JC1 Lecture Test 1 (Mole Concept and Acid-Base Titration) v8
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Temasek Junior College 2012 Chemistry Lecture Test 1 Mole Concept This Test consists of 2 printed pages: Answer ALL the questions: 1. 10 cm 3 of a gaseous organic compound, C x H y O was exploded with 100 cm 3 of oxygen. The volume of the gaseous mixture decreased to 90 cm 3 . A further reduction of 30 cm 3 was observed when the residual gases were passed through aqueous sodium hydroxide. (a) Write an equation for the combustion of the organic compound [1] (b) Deduce the molecular formula of the organic compound. [3] (c) In another run of the same experiment, 40 cm 3 of oxygen was used instead. It was observed that the contraction in volume after the reaction was smaller in magnitude. Also, when the product mixture was passed through sodium hydroxide, the contraction in volume was smaller in magnitude. Suggest a plausible explanation for this. [1] 2 Hydrogen peroxide is a strong oxidising agent that can be used as a disinfectant in swimming pools. To ensure that there is no algae or bacterial growth, the minimum concentration of hydrogen peroxide must be at least 40 ppm. Hydrogen peroxide is commercially available as a solution with concentration 8.82 mol dm -3 . (a) Calculate the minimum volume of commercially available hydrogen peroxide needed to disinfect an Olympic-sized swimming pool containing 3000 m 3 of water. (You may take the density of the pool water as 1 g cm -3 ) [2] (b) Hydrogen peroxide decomposes into water and oxygen when exposed to sunlight. 1
Transcript of JC1 Lecture Test 1 (Mole Concept and Acid-Base Titration) v8
Temasek Junior College 2012 Chemistry Lecture Test 1Mole Concept
This Test consists of 2 printed pages:
Answer ALL the questions:
1. 10 cm3 of a gaseous organic compound, CxHyO was exploded with 100 cm3 of oxygen. The volume of the gaseous mixture decreased to 90 cm3. A further reduction of 30 cm3 was observed when the residual gases were passed through aqueous sodium hydroxide.
(a) Write an equation for the combustion of the organic compound [1]
(b) Deduce the molecular formula of the organic compound. [3]
(c) In another run of the same experiment, 40 cm3 of oxygen was used instead. It was observed that the contraction in volume after the reaction was smaller in magnitude. Also, when the product mixture was passed through sodium hydroxide, the contraction in volume was smaller in magnitude. Suggest a plausible explanation for this. [1]
2 Hydrogen peroxide is a strong oxidising agent that can be used as a disinfectant in swimming pools. To ensure that there is no algae or bacterial growth, the minimum concentration of hydrogen peroxide must be at least 40 ppm.
Hydrogen peroxide is commercially available as a solution with concentration 8.82 mol dm-3.
(a) Calculate the minimum volume of commercially available hydrogen peroxide needed to disinfect an Olympic-sized swimming pool containing 3000 m3 of water. (You may take the density of the pool water as 1 g cm-3) [2]
(b) Hydrogen peroxide decomposes into water and oxygen when exposed to sunlight.
(i) Write a balanced equation for hydrogen peroxide decomposing to form oxygen and water.
(ii) Calculate the volume of oxygen formed at room temperature when all the hydrogen peroxide in the swimming pool in (a) is decomposed. [2]
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3 Solution A is an aqueous alkali solution of Ba(OH)2 formed by dissolving a certain mass of Ba(OH)2 in 250 cm3 of water.
Solution B is formed by diluting 60 cm3 of solution A with 20 cm3 of water. It was found that 12 cm3 of solution B required 18 cm3 of 0.3 moldm-3 H2SO4 for complete reaction.
(a) (i) Find the concentration of solutions A and B.
(ii) Hence, find the mass of Ba(OH)2 dissolved to form solution A. [4]
(b) Solution C is formed by mixing 35 cm3 of the above solution A with 70 cm3 of solution B. It was found that 30 cm3 of solution C required 25 cm3 of H3PO4 for complete reaction.
(i) Write a balanced equation for the reaction between Ba(OH)2 and H3PO4.
(ii) Find the concentration of the H3PO4 reacted. [4]
4 Sulfuric acid is a mineral acid which is commonly used in analytical chemistry for the titration of various bases. Unlike organic acids, it is soluble in water at all concentrations and is a strong acid.
50 cm3 of 0.3 mol dm-3 potassium hydroxide is placed in a conical flask and sulfuric acid is gradually added to it.
(a) Give the balanced equation for the reaction that takes place. [1]
(b) Given that the concentration of the sulfuric acid is 0.1 mol dm-3, what is the minimum volume of sulfuric acid to be added such that the concentration of the salt formed is 0.05 mol dm-3. [3]
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1. 10 cm3 of a gaseous organic compound, CxHyO was exploded with 100 cm3 of oxygen. The volume of the gaseous mixture decreased to 90 cm3. A further reduction of 30 cm3 was observed when the residual gases were passed through aqueous sodium hydroxide.
(a) Write an equation for the combustion of the organic compound [1]
CxHyO(g) + (x + – ) O2(g) à x CO2(g) + H2O(l)
(b) Deduce the molecular formula of the organic compound. [3]
CO2 reacts with NaOH. Hence volume of CO2 gas = 30 cm3
● Volume of oxygen remaining after reaction = 90 – 30 = 60 cm3
Volume of oxygen reacted = 100 – 60 = 40 cm3
OR
CxHyO(g) + (x + – ) O2(g) à x CO2(g) + H2O(l)
Initial vol 10 100 0 -Change in vol -10 -40 +30 -Final vol 0 60 30 -
For CxHyO : CO2
● x = 3
For CxHyO : O2
From the equation,
● y = 6
The carbonyl is C3H6O.
(c) In another run of the same experiment, 40 cm3 of oxygen was used instead. It was observed that the contraction in volume after the reaction was smaller in magnitude. Also, when the product mixture was passed through sodium hydroxide, the contraction in volume was smaller in magnitude. Suggest a plausible explanation for this. [1]
● There is incomplete combustion, hence the oxygen used up was less than expected, and less carbon dioxide was produced.
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2. Hydrogen peroxide is a strong oxidising agent that can be used as a disinfectant in swimming pools. To ensure that there is no algae or bacterial growth, the minimum concentration of hydrogen peroxide must be at least 40 ppm.
Hydrogen peroxide is commercially available as a solution with concentration 8.82 mol dm-3.
(a) Calculate the minimum volume of commercially available hydrogen peroxide needed to disinfect an Olympic-sized swimming pool containing 3000 m3 of water. (You may take the density of the pool water as 1 g cm-3) [2]
Concentration of H2O2 required = 40 ppmThere is 40 g of H2O2 in 1,000,000 g pool water
Given that the density of the pool water = 1 g cm-3
There is 40 g of H2O2 in 1,000,000 cm3 = 1000 dm3 of pool waterVolume of pool water = 3000 m3 = 3000000 dm3
● Mass of H2O2 needed for disinfection =
= 120000 g
Amount of H2O2 needed for disinfection =
= 3530 mol
● Volume of commercially available H2O2 needed =
= 400 dm3
(b) Hydrogen peroxide decomposes into water and oxygen when exposed to sunlight.
(i) Write a balanced equation for hydrogen peroxide decomposing to form oxygen and water.
● 2H2O2 → O2 + 2H2O
(ii) Calculate the volume of oxygen formed at room temperature when all the hydrogen peroxide in the swimming pool in (a) is decomposed. [2]
Amount of H2O2 in the swimming pool = 3530 mol
● Volume of O2 formed =
= 42360 dm3 = 42400 dm3 (3sf)
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3 Solution A is an aqueous alkali solution of Ba(OH)2 formed by dissolving a certain mass of Ba(OH)2 in 250 cm3 of water.
Solution B is formed by diluting 60 cm3 of solution A with 20 cm3 of water. It was found that 12 cm3 of solution B required 18 cm3 of 0.3 moldm-3 H2SO4 for complete reaction.
(i) Find the concentration of solutions A and B.
Ba(OH)2 + H2SO4 à BaSO4 + 2H2O
Ba(OH)2 H2SO4
Number of moles of H2SO4 reacted =
= 0.00540 mol (3sf)
Number. of moles of Ba(OH)2 in 12cm3 of B = 0.00540 mol
Concentration of solution B =
= 0.45 moldm-3
Using MAVA = MBVB
Where MA = Conc. of solution AVA = Vol. of solution AMB = Conc. of solution BVB = Vol. of solution B
Concentration of solution A =
= 0.60 moldm-3
(ii) Hence, find the mass of Ba(OH)2 dissolved to form solution A. [4]
Number of moles of Ba(OH)2 in 250 cm3 of A =
= 0.15 mol
Mass of Ba(OH)2 dissolved = 0.15 x [137 + 2 x (16 + 1)]= 25.7 g
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(b) Solution C is formed by mixing 35 cm3 of the above solution A with 70 cm3 of solution B. It was found that 30 cm3 of solution C required 25 cm3 of H3PO4 for complete reaction.
(i) Write a balanced equation for the reaction between Ba(OH)2 and H3PO4.
3Ba(OH)2 + 2H3PO4 à Ba3(PO4)2 + 6H2O
(ii) Find the concentration of the H3PO4 reacted.
Total number of moles of Ba(OH)2 in solution C formed
=
= 0.0525 mol
Total volume of solution C formed = 35 + 70= 105 cm3
Concentration of solution C =
= 0.50 moldm-3
3 Ba(OH)2 2 H3PO4
Number of moles of Ba(OH)2 reacted =
= 0.0150 mol
Number of moles of H3PO4 that reacted =
= 0.0100 mol
Concentration of H3PO3 =
= 0.40 moldm-3
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4. Sulfuric acid is a mineral acid which is commonly used in analytical chemistry for the titration of various bases. Unlike organic acids, it is soluble in water at all concentrations and is a strong acid.
50 cm3 of 0.3 mol dm-3 potassium hydroxide is placed in a conical flask and sulfuric acid is gradually added to it.
(a) Give the balanced equation for the reaction that takes place. [1]
● 2KOH + H2SO4 → K2SO4 + 2H2O
(b) Given that the concentration of the sulfuric acid is 0.1 mol dm-3, what is the minimum volume of sulfuric acid to be added such that the concentration of the salt formed is 0.05 mol dm-3. [3]
Let the volume of the sulfuric acid be V cm3
Since the volume of H2SO4 to be used is a minimum, we can assume sodium hydroxide is in excess.
● No. of moles of salt formed (K2SO4) = 1000
1.0V
= 1000
1.0 V
● Concentration of Na2SO4 =
=
= 0.05 moldm-3
Solving,0.1V = 2.5 + 0.05V
V = 50 cm3
Note:
No. of moles of KOH used = = 0.006
Which is less than the total amount of KOH present KOH is in excess
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