# UTME Chemistry JAMB Past Questions and Answers, Jamb, Pre-degree, Nursing, Waec, Neco

• How many moles of oxygen molecules would be produced from the decomposition of 2.50 moles of potassium trioxochlorate (V)?
a. 2.50 b. 3.50 c. 3.75 d. 7.50
SOLUTION
2KClO3(s) → 2KCl(s) + 3O2(g)
If 2moles of KCLO3 gives 3 moles of oxygen (O2)
2.5 moles will give Xmoles
Xmoles = 2.5 × 3/2
= 3.75 mole
• : Na2CO3 + 2HCL 2NaCL + H2O + CO2(g). Using the above equation, what volume of carbondioxide, measured at s.t.p is liberated when 53g of sodium carbonate is dissolved in hydrochloric acid? ( 1 mole of gas occupies 22.4dm3 at s.t.p) (Na = 23, C = 12, O = 16)
a. 44.8 dm3 b. 11.2 dm3 c. 100.1 d. 3.0 dm3 e. 22.4 dm3
SOLUTION
Na2CO3 + 2HCL → 2NaCL + H2O + CO2(g).
Molar mass of Na2CO3 = 23×2 + 12 + 16×3
46+12+48 = 106gmol–¹
No. Of moles = mass/molar mass
53/106 = 0.5moles ( Na2CO3 + 2HCL gives one mole of CO2)
No. Of moles = volume at s.t.p/ molar volume;
Volume at S.t.p = 0.5 × 22.4
11.2 dm³. [ B ]
• Given that 32.0g sulphur contains 6.02 × 1023 sulphur atoms. How many atoms are there in 2.70g of alluminium? (AL = 27, S = 32)
a. 6.02× 1023 b. 3.01× 1022 c. 6.02 × 1022 d. 5.08× 1022 3.01× 1022
SOLUTION
No. Of moles of aluminum = 2.7/27; = 0.1 moles
No. Of atoms = No. Of moles × Avogadro’s constant
0.1 × 6.02 × 10²³. = 6.02 × 10²². [ C]
• If 10cm3 of distilled water is added to 10cm3 of an aqueous salt solution, the concentration of the solution
a. increases b. decreases c. remains constant d. doubles
SOLUTION
Addition of water to a solution decreases the level of concentration of the solution, this is called a diluted solution. NOTE: if you did not add water or any other solvent and you take a particular volume out of the solution, the concentration remains the same (imagine you have a cup full of water and you dissolved a tea spoon full of sugar on it. If you take a quantity out of the cup, does the solution taste any difference? No!). Changing the amount of solute or solvent is what affects concentration of a solution. ANS:B
• Which of the following quantities is the same for one mole of Br2(l) and one mole of He(e)?
a. number of molecules b. number of atoms c. total mass of the atoms d. volume occupied at s.t.p
SOLUTION
Their number of molecules will be the same for one mole. They both contain 6.02 × 10²³ molecules. But their number of atoms is different because Br2 is diatomic (2 × 6.02 × 10²³) while He is monoatomic (6.02×10²³). Since they have different atomic masses, their total mass cannot be the same. They could have occupied the same volume since they have same number of molecules, but Br2 is a liquid, the rule only applies to gases.
ANS: [ A ]
• During a titration experiment, 0.05 moles of carbon (IV) oxide is liberated. What is the volume of CO2 gas liberated?
a. 22.40 dm3 b. 11.20 dm3 c. 2.24 dm3 d. 1.12 dm3
SOLUTION
Volume of CO2 = 22.4 × 0.05 ; 1.12dm³ [D]
• If a solution contains 4.9g of tetraoxosulphate(vi) acid, calculate the amount of copper (II) oxide that will react with it [Cu = 64, O = 16, S = 32, H = 1]
a. 0.8g b. 4.0g c. 40.0g d. 80.0g
SOLUTION
Molar mass of H2SO4 = 1×2 + 32 + 16×4; = 2+32+64
= 98mol–¹
First let’s find the number of moles of tetraoxosulphate(vi) acid,
No. Of moles = mass/molar mass
4.9/98 = 0.05moles
Molar mass of CuO = 64 + 16; = 96gmol–¹
Mass = 0.05×80; = 4.0g
• Calculate the volume of carbon (IV) oxide measured at s.t.p, produced when 1kg of potassium hydrogen trioxocarbonate (IV) is totally decomposed by heat? (G.M.V at s.t.p = 22.4 dm3 , K = 39, O = 16, C = 12, H = 1)
a. 28 dm3 b. 56 dm3 c. 112 dm3 d. 196 dm3
SOLUTION
2KHCO3(s) → K2CO3 (s) + H2O(g) + CO2(g)
Volume of CO2 = ?
Molar mass of KHCO3 = 39+1+12+16×3; = 100gmol–¹, mass = 1000g
No. Of moles of potassium hydrogen trioxocarbonate (IV), KHCO3 = 1000/100
= 10moles
The mole ratio of KHCO3 to CO2 = 2:1
Therefore, No. of moles = 10/2 = 5moles
Volume of CO2 = No. of moles × molar volume
5 × 22.4 = 112dm³
• If 200cm3 of a gas at s.t.p has a mass of 0.268g, what is its molar mass? (Molar volume of a gas at s.t.p = 22.4 dm3 )
a.300gmol-1 b 200 gmol-1 c. 150 gmol-1 d. 30 gmol-1 e. 15 gmol-1
SOLUTION
No. Of moles = volume/molar volume ; = 200/22.4 ÷ 1000 ( converting cm to dm)
8.923/1000 = 0.008923moles
Molar mass = mass/No. of moles; = 0.268/0.008923
= 30gmol–¹
• What volume of oxygen at s.t.p would react with carbon to form 4.40g of CO2 according to the following equation? C(s) + O2(g) CO2(g) [ o = 16, c = 12; 1 mole of a gas occupies 22.4dm3 at s.t.p]
a. 0.224 dm3 b. 2.24 dm3 c. 4.40 dm3 d. 4.48 dm3 e. 22.4 dm3
SOLUTION
CO2 → O2 + C
C(s) + O2(g) → CO2(g)
The mole ratio = 1:1
Molar mass of CO2 = 12 + 16×2; = 44gmol–¹
No. of moles of CO2 = mass/molar mass; = 4.40/44; = 0.1moles
Volume of oxygen = molar volume × No. of moles
22.4×0.1 = 2.24dm³. [B]
• If 1 mole of sodium contains 6 × 1023 atoms, how many atom are contained in 0.6g of sodium? [Na = 23, ]
a. 1.56 × 1023 b. 1.56 × 1022 c. 3.6 × 1023 d. 3.6× 1022
SOLUTION
No. of atoms = No. of moles × Avogadro’s constant
No. of moles = 0.6/23; = 0.026moles
0.026×6.02×10²³. = 1.56 × 10²² [B]
• If 20cm3 of distilled water is added to 80cm3 of 0.50 moldm-3 hydrochloric acid, the new concentration of the acid will be
a. 0.10 moldm-3 b. 0.20 moldm-3 c. 0.40 moldm-3 d. 2.00 moldm-3
SOLUTION
2HCl +2H2O → H2O2 + 2H2 + Cl2
VA = 20cm³, Ca = ?, Vb = 80cm³, Cb = 0.50 moldm–³, Na/Nb =
CaVa/CbVb = Na/Nb
20×Ca/0.50×80 = 1/1
Ca = 40/20; = 2.00moldm–³
• Consider the reaction represented by the equation. 2Na2HCO3(s) 2Na2CO3(s) + CO2(g) + H2O(g). What volume of carbon (IV) oxide at s.t.p is evolved when 0.5 moles of Na2HCO3 is heated? [molar volume = 22.4dm3 at s.t.p]
a. 1.12 dm3 b. 2.24 dm3 c. 5.6 dm3 d. 56.0 dm3
SOLUTION
2Na2HCO3(s) → 2Na2CO3(s) + CO2(g) + H2O(g).
2 moles of Na2HCO3 gives 1 mole of CO2; therefore 0.5/2 = 0.25moles
Volume = 0.25 × 22.5; = 5.6dm³
• An iron ore is known to contain 70.0% Fe2O3 . The iron obtained from 80kg of the ore is? (Fe = 56, O = 16]
a. 35.0kg b. 39.2kg c. 70.0kg d. 78.4kg
SOLUTION
Ore is defined as a mineral that contains a metal of interest. We will talk more about ores when we get to metals and non-metals. Firstly, let is know the amount of Fe2O3 in the ore;
%Fe2O3 = mass of Fe2O3 /mass of ore × 100
70 = mass of Fe2O2/80 × 100
Mass of Fe2O3 = 56kg
The decomposition of Fe2O3 gives
2Fe2O3→ 4Fe + 3O2
Molar mass of Fe2O3 = 56×2 + 16×3; = 160gmol–¹
Number of mole of Fe2O3 = mass/molar mass; = 56×10³/160
= 350 moles
The mole ratio of Fe2O3 to Fe is 2:4 which gives 1:2
Number of mole of Fe = 350×2; = 700moles
Mass of Fe = No. Of moles × molar mass
700×56 = 39200; = 39.2kg
• In two separate experiments, 0.36g and 0.71g of chlorine combined with a metal X to give Y and Z respectively. An analysis showed that Y and Z contain 0.20g and 0.40g of X respectively. The data above represents the law of?
A. Multiple proportion. B. Conservation of mass. C. Constant composition.
D. Reciprocal proportion
SOLUTION
The ratio of chlorine to that of X (1:2) in the two compounds Y and Z remains the same. This obeys the law of constant composition or definite proportion.

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