1987 JAMB CHEMISTRY ANSWERS

UTME 1987 CHEMISTRY ANSWERS

1.     An ink is a mixture of coloured ions and a mixture of coloured substances can be separated into their various components by chromatography. Hence, the ions: Co^2+, Cu²⁺ and F2²⁺ present in an ink can best be separated by chromatography [D]

2.     A mixture consists of two or more constituents which can be separated by physical method. Components of granulated sugar, sodium chloride and iron filings cannot be separated by physical method while sea water can be separated into its component of salt and water by evaporation

CaCO3 + 2HCl  CaCl2 + H2O + CO2

3.     From the above equation:

100g of CaCO3 produced 1 mole of CO2

10g of CaCO3 will produce 0.1 mole of CO2

1 mole of CO2 contains 6.02 1023 molecules

0.1 mole of CO2 will contain   ×  6.02 ×  10²³

= 6.02 × 10²²  [C]

4.     CaC₂ + 2H₂O  Ca(OH)₂  +  C₂H₂

From the above equation:

64g of CaC2 produced 1 mole of acetylene (ethyne). 1 mole of gas at s.t.p occupies GMV of 22400cm3

If 22400cm3 of C2H2 was produced by 64g of CaC2, 1000cm3 of C2H2 will require

  ×     =  2.9g  [B]

5.     It is very important to note that in Boyle's law, the volume of a gas is NOT the same as the quantity of a gas but the space occupied by the gas. If the quantity of O2 in a given container is reduced by one-half of the quantity of oxygen will occupy double the original space ie V2 = 2V1

Thus, V2 = 2 ×  2.76 = 5.52litres

Applying Boyle's law: P1V1 = P2V2

0.825  2.76 = 5.52P2

P2 =   = 0.413 atm  [C]

6.     Molecular mass of ethanoic acid (CH3COOH) = 60g

Molecular mass of propanol (C3H7OH) = 85g

Molecular mass of ethanal (CH3CHO) = 44g

Since vapour density is directly proportional

to molecular mass. Thus, the substance with the

lowest molecular mass has the lowest vapour

density  [D]

7.     According to Graham's law of diffusion, the rate of diffusion of a given mass of a gas is inversely proportional to the square root of its density at constant temperature and pressure This is expressed mathematically as:

R       =  R =     [A]

8.     Atomic number = number of protons = 17.

Mass number = number of protons (P) + number of neutrons (N)

N = M - P. N = 36 -  17 = 19

Thus, the number of neutrons and protons are 19 and 17  [C]

9.     Electronic configuration of X = 2, 8, 2. The number of valence electrons in X = 2, hence X belongs to group II in the periodic table. Electronic configuration of Y: 2, 7. The no of valence electrons in Y = 7, Hence Y belongs to group VII in the periodic table. Elements that are further away in the periodic table. Elements that are further away in the periodic table or that have a wide electronegativity difference of ≥ 1.67 form an ionic bond  [A]

10.  Contribution of  =   ×     = 14.4

Contribution of  =    ×    = 1.8

Relative atomic mass of Z = (14.4 + 1.8) = 16.2g [B]

11.  The greater the difference in the electronegativity of atoms of 2 elements, the greater their tendency to form an ionic bond and ionic compounds exhibit high polarity. [B]

12.  Air is a mixture of gases namely O2, N2, CO2, water vapour and inert gases. When passed through tube Y, red hot Cu absorbs the O2. When passed through tube Z, CaCl2 absorbs the water vapour. Hence, the remaining gases are N2 and the inert gases  [C]

13.  In water purification, alum like potash alum KAl(SO4)2 or sodium aluminate (III) NaAlO2 are added to water to cause clumping together of small particles of dirt . The process is called coagulation or flocculation [D]

14.  Hard water when titrated with soap has higher titre value than soft water. Temporary hard water after boiling becomes soft, rain water is a soft water, permanent hard water passed through permutit becomes soft. However, permanent hard water after boiling still retains its hardness  [A]

15.  G5

16.  Mass of NaAsO4(H2O)12 contained in 100g of H2O = 38.9g

Molar mass of Na3AsO4 (H2O) = 424g

Molar mass of Na3AsO4 = 208g

Mass of Na2AsO4 contained in 100g of H2O

 =    ×       = 19g

 Mass of saturated solution of NaAsO4(H2O)12 = (38.9 + 100) = 138.9g

% composition of Na3AsO4 in the saturated solution =    ×    = 13.7%

17.  Lime juice contains citric acid and thus exhibit the properties of an acid which include : liberates CO2 gas with NaHCO3, forms red colour with methylorange, has a sour taste and liberates H2 gas with sodium [A]

18.  Hydrochloric acid is a strong acid, ethanoic acid is a weak acid, sodium chloride is a neutral salt, milk of magnesia Mg(OH)2 is a weak base, Sodium hydroxide is a strong base  [C]

19.  The formula of tetraoxophosphate (V) acid is H3PO4. This ionizes as follows: H3PO4   3H+  + PO43-

The number of replaceable H+ = 3. Basicity of H3PO4 = 3  [D]

20.  Equation for the reaction:

NaOH + HCl  NaCl  +  H2O

Mole ratio of acid and base     =  

Using the volumetric formula:   =   

  =    Ma = 0.094  [A]

21.  Electrolytic equation for silver liberation:

Ag+  + e-  Ag

1 mole of Ag+ requires 1F.

Electrolytic equation for liberation of Al:

Al3+  + 3e-  Al

1 mole of Al requires 3F

3.6g of Ag =   = 0.03mole

0.03mole of Ag requires 0.03F

Number of mole of Al that will be liberated by 0.03F =  0.03F =  ×   = 0.01mole

0.01mole of Al = 0.01 × 27 = 0.27   0.3g  [D]

22.  If 1 Faraday of electricity is passed through 1M CuSo4 for 1 minute, 60 moles of Cu will be liberated at the cathode in preference to H. OH - will be discharged at the anode in preference to SO42-. H+ interacts with SO42- in the solution to form an acid which decreases the pH of the solution at the anode [B]

23.  Electrolytic equation for Mg ion:

Mg2+ + 2e-  Mg

From the above equation, 1 mole (24g) of Mg requires 2F = 2 × 96500C of electricity.

Quantity of electricity passed = (2 × 900) = 18000C

Mass of Mg that would be obtained =  ×    = 2.24g  [C]

24.  Half ionic equation for Cu from the given equation: 3Cu2=  + 6e-  3Cu

Half ionic equation for 1 mole of Cu:

Cu2+  + 2e-  Cu

Number of electrons transferred = 2

1 mole of electron contains Avogadros number = 6.02 ×  10²³. 2 moles of electron = 2 × 6.02 × 10²³ = 1.2 × 10²⁴  [C]

25.  Both sodium hydrogen trioxocarbonate (IV) and iron (II) trioxocarbonate (IV) (FeCO3) would liberate CO2 with acids. However, only iron (II) trioxocarbonate can reduce  (decolourize) the acidified KMnO4 and itself being oxidized to iron (III) trioxocarbonate  [C]

26.  Amount of heat absorbed by 5g of NH4NO3 when dissolved in water = 1.6KJ

Molecular mass of NH4NO3 = 80g

Heat of solution of NH4NO3 =   ×    = 25.6 [B]

27.  Product = H2SO4, reactant = SO3 and H2O

Enthalpy change  H = Hp - HR

= (-811) - (-395 + - 286)

-811 - (- 681)

= -811  +  681 = -130KJ

28.  From the given table, the temperature row shows increase in temperature by 100C ie (35 - 25) = 10 and (45 -35) = 10.

The time row shows a decrease in time by half ie:    =     and    =  . Since time is inversely related to rate, the time row shows an increase in the rate of reaction by 2   [A]

29.  In the given reaction, the enthalpy change H is negative .This shows that the forward reaction is exothermic. In exothermic reaction, a decrease in temperature leads to increased formation of the product (SO3)  [C]