Tuesday, 7 March 2017

1983 JAMB CHEMISTRY ANSWERS



UME 1983 CHEMISTRY ANSWERS
1.     If a solution of X turns litmus red, this suggests that X must be acidic. Among the options only NaHCO3 and NaHSO4 are acidic. However, NaHCO3 does not react with Na2CO3 to liberate CO2 that turns lime water milky. Only NaHSO4 can liberate CO2 with Na2CO3 and forms a white precipitate with BaCl2 that is ionsoluble in dilute HCl [C]
2.     The products of hydrolysis of fats and oil with caustic soda  (saponification reaction) are sodium octadecanoate (soap) and propane 1,2,3-triol (glycerol). Thus, the alkanol is glycerol [B]
3.     Flames that can produce heat of a very high temperature (over 20000C) are used by welders in cutting metals. Such flames are oxy-acetylene(oxy-ethyne) flame and oxy-hydrogen flame [D]
4.     Homologous series is a family of organic compound  which follows a regular structural pattern in which each successive member differ in its molecular formula by CH2 group [B]
5.     From the electronic configuration: 2, 8, 6, the valence electron is 6. This means that the element belongs to group VI in the periodic table. S-block elements are group I and II elements while P-block elements are group III – VIII elements. Thus the elements is a P-block element [D]
6.     Let the number of water molecules be x
Weight of the hydrated salt (CuSO4.5H2O)
= (14.98 - 10) = 4.98g
Weight of the dehydrated salt (CuSO4) = 159.5g
Molecular mass of water = 18g
 =
794.3
63.7x = 229.7
X =  = 3.6  4 [D]
7.     Hybridization of methane forms a new set of four identical hybrid orbitals called sp3 by transferring one electron from 2S orbital to 2Pz orbital as shown below:

1S
2S
2Px
2Py
2Pz
2Pz
2Py
2Px
2S
1S
                        









The four unpaired electrons in the central carbon atom become covalently bonded with 4-H atoms located at the apices to give rise to a tetrahedral structure [D]
8.     A compound with the elements C. H and O is a carbohydrate. Among carbohydrates, glucose has the empirical formula of CH2O. It is also a colourless crystalline solid, sweet to taste , melts on heating and readily soluble in water. Thus X is glucose. In the presence of yeast, glucose is converted to ethanol and CO2 which is a colourless gas. Hence Y is ethanol. Ethanol reacts with ethanoic acid to form an ester called ethyl ethanoate which is the sweet smelling compound [D]
9.     Empirical formula of X = CH2O
Molecular mass of X = 180g
(CH2O)n = 180
(12+ 2 + 16)n = 180
30n = 180
n =  = 6
Molecular formula = (CH2O)6 = C6H12O6 [B]
10.  The reaction of X which is glucose with yeast to produce carbon (IV) oxide and ethanol is called fermentation. Fermentation is widely applied in brewing industries for the production of alcoholic drinks [C]
11.  A mixture of NaCl, NH4Cl and BaSO4 should first be heated to convert the solid NH4Cl into gaseous NH3 and HCl. This process is called sublimation. Water should then be added to the remaining mixture to form a solution of soluble NaCl and insoluble BaSO4. The insoluble BaSO4 is then filtered off.
12.  The ideal gas equation which shows the behaviour of ideal gases is a combination of the three gas laws namely Charles law Boyles law and pressure law. This states that pressure is directly proportional to temperature and inversely proportional to volume.
This is expressed mathematically as follows:
P         P  =       = K
  =      [B]
13.  Heating  solid NH4Cl converts it into gaseous products namely NH3 and HCl. Since NH3 is lighter than HCl, it will diffuse faster to reach the damp neutral litmus paper and being an alkaline gas turns it blue before the arrival of the acidic HCl which will introduce a tinge of redness into the paper giving rise to a final purple colour which is a mixture of blue and  red colours.
14.  The flame tests for cations are as follows:

CATIONS
FLAME COLOUR
K+
Lilac
Na+
Golden yellow
Ca2+
Brick-red
Ba2+
Pale green
Cu2+
Deep green
Pb2+
Deep green
Sr2+
Bright red              [C]

15.  From the equation of chemical equilibrium, the enthalpy change ( H) is positive which indicates an endothermic reaction. Increasing the temperature of endothermic reaction favours the forward reaction leading to increased concentration of the product. [E]
16.  In equation (I):  the metallic iron (Fe) changed to iron (II) tetraoxosulphate (VI) (FeSO4). Change in oxidation number = 0 to +2. Thus iron is oxidized. In equation (III) : iron (II) chloride  changed to iron (III) chloride. Change in oxidation number =  +2 to +3. Thus, iron is oxidized [D]
17.  Electrolytic equation for the reduction:
Cu2+ + 2e-   Cu
Ag+ e-   Ag
From the above equation, the quantity of electricity that produces 1 mole of Cu will produce 2 moles of Ag.
Mass of Cu deposited = 0.63 =  = 0.01 mole
No of mole of Ag that will deposit  = 0.02 mole
Mass of 0.02 mole of Ag = (0.02 × 108) = 2.16g [C]
18.  In the reaction between a metallic atom and a metallic ion, the more electropositive metal will displace the less electropositive one from its solution. Elements that are higher in the electrochemical series have higher electropositivity [E]
19.  The presence of double bond at C-1 atom indicates series alkene named as (1-ene). The longest carbon chain has 4 C-atoms named as (but). The branch chain is methyl group at C-2 atom named as (2- methyl). Combining the names in the reverse order gives the IUPAC name 2-methylbut-1-ene [C]
20.  Isomers of dibromopropane (C3H6Br2) are as follows: 1,1- dibromopropane, 2,2 dibromopropane and 1, 3- dibromopropane [D]
21.  Suphur reacts with oxygen to form SO2 which is an acid anhydride that readily dissolves in water to form trioxosulphate (IV) acid (H2SO3) [C]
22.  The process whereby a substance loses its water loses its water of crystallization on exposure to atmosphere to form a lower dehydrate or anhydrous compound is called effloresecence eg: Glauber’s salt (Na2SO4.10H2O), washing soda (Na2CO3. 10H2O), Epsom salt (MgSO4.7H2O) green vitriol (FeSO4.7H2O), white vitriol (ZnSO4.7H2O), etc. [A]
23.  Molten sodium chloride consists of two ions namely Na+ and Cl-. During electrolysis, Na+ migrate towards the cathode where it gains an electron and become discharged as Na while Cl-1 migrate to the anode where it loses an electron and become discharged as Cl. The Cl- which has lost electron to form a chlorine atom is said to have been oxidized
24.  Crude petroleum pollutant in the creeks and waterways can be dispersed using dispersant. The ideal dispersant used is detergent. When detergent is introduced into a crude oil polluted water body, the hydrophilic head dissolves in water while the hydrophobic tail dissolves in the oil. The oil then floats off as globule surrounded by detergent molecules. The oil which is now dispersed is then emulsified by the mechanical action of the water current thereby removing it.
25.  Electronegativity is the power of an atom in a molecule to attract an electron to form a mole of gaseous ion with a negative charge. Thus, an element  is said to be electronegative if it has this tendency.
26.  According to the pH  scale, acidity decreases with increase pH. Alkalinity increases with increase in pH  and decreases with a decrease in pH . Solution Z haing the highest pH  is the most alkaline and hence less acidic  [E]
27.  In the given equations (i) and (ii) the enthalpy changes (H) are both negative. This means that both reactions are exothermic and heat energy is evolved. However, neglecting the negative signs, the numerical value of H in equation(I) Is smaller than H in equation (II). This means that less heat is evolved in equation (I) [C]
28.  Among the elements given, only Cu is below hydrogen in the electrochemical series and thus cannot displace H from its solution. Mg, Fe and Pb are above  H in the electrochemical series. However, Pb cannot displace H from its solution due to the formation of insoluble outer layer of lead (II) chloride. Hence, only Mg and Fe can displace H from its solution [D]
29.  Ideally, stainless steel is an alloy formed by mixing the base metal , iron with little amount of chromium and nickel (Fe, Cr & Ni). However, in the absence of nickel, carbon can be used . Thus, Fe, C, & Cr is also correct. [B]
30.  Equation for the reaction: H2SO4 + 2NaOH  Na2SO4 + 2H2O
Mole ratio of acid to base  =
Applying the volumetric formula:  =
Substitute:
Va = 2cm3 [A]
31.   


32.  Highly electropositive elements such as K, Na, Ca Mg and Al are extracted by electrolysis. Less electropositive elements such as Fe, Cu, Sn and Pb are mainly extracted by thermal reaction. In addition, Al ore (bauxite) first undergoes some thermal reactions to yield the pure aluminate (Al2O3) which is then electrolysed (TL. Cu after thermal reaction is purified by electrolysis (TL)) Fe is extracted ONLY by thermal reaction [N.C.O]
33.  Cu(NO3)2 is prepared by the action of Cu on concentrated trioxonitrate (V) acid. When prepared from CuO, the first step is to reduce the heated CuO to metallic Cu using CO or H. Reduction of CuO cannot be achieved by reacting CuO with H2SO4 as in option A [A]
34.  Most alcoholic drinks including palm wine contains less than 95% ethanol. To convert them to high quality ethanol with a concentration greater than 95%, the wine would be subjected to fractional distillation with CaO as a dehydrant [C]
35.  From Boyle’s law, at constant temperature, volume of a given mass of a gas is inversely proportional to its pressure. At constant mass, volume is inversely proportional to density. Hence, increase in pressure causes a decrease in volume which in turn causes an increase in density [D]
36.  Mass of the hydrated salt = 2.5g
Mass of anhydrous salt = 2.13g
Molecular mass of anhydrous salt = 208g
Let the number of water molecules be x
Molecular mass of water present = 18x
 =
38.34X + 443.04 = 520
38.34x = 520 – 443.04 = 76.96
X =  = 2 [D]
37.  Mass of KClO3 that saturated 10cm3 of water = 3.06g
Mass of KClO3 that will saturate 1dm3 (1000cm3) of water =  ×  = 306g
Molecular mass of KClO3 = 122.5g
Solubility of KClO3 =  = 2.5moldm-3 [C]
38.  Large molecular hydrocarbons such as diesel, kerosene, gas oil, etc are broken down into smaller molecular hydrocarbon like petrol which is used as fuel for many car engines [C]
39.  SO2 is a strong reducing agent in the presence of water. It reduces chlorine to hydrochloric acid. However, in the presence of a stronger reducing agent like H2S it acts as an oxidizing agent and oxidizes H2S to a solid sulphur.
40.  All tetraoxosulphates salts give white precipitate with acidified BaCl solution. In addition, flme test shows that the colour of Cu flame is green [B]
41.  The masss number of an atom of an element is defined as the sum of the number of protons and neutrons in the nucleus of an atom [D]
42.  Neutralization reaction involves the formation of salt and water from the reaction of acid and base. In option E, the reactants are HNO3 which is an acid and KOH which is a base. Both reacts to form salt (KNO3) and water (H2O) as shownas follows: HNO3 + KOH  KNO3 + H2O
43.  General equation for combustion of hydrocarbons
CxHy + ( x + )O2   XCO2 +  H2O
Replace x with 2 and y with 6 to form the equation for combustion of ethane:
C2H6 + (2 +  ) O2   2CO2 + 3H2O
C2H6 + 3  O2   2CO2 + 3H2O
From the above equation:
30g of ethane produced 54g of water
300g of ethane will produce
  = 5400kg of water
Thus, 3000kg of ethane was replaced by the heavier 5400kg of water.
Gain in weight = (5400 - 3000) = 2400kg [D]
44.  One of the chemical properties of an acid is the ability to react with trioxocarbonates to produce CO2 which turns lime water Ca(OH)2 milky [D]
45.  Sodium metal dissolves in water to produce sodium hydroxide and hydrogen and NOT oxygen .The equation is shown below:
2Na + 2H2O  2NaOH + H2 [B]
46.  When the enthalpy change (H) is positive, the reaction is said to be endothermic because heat energy is absorbed (gained) from the surroundings [B]
47.  From the equation of the equilibrium reaction, the enthalpy change is negative (-) showing that the reaction is exothermic . The forward reaction of an exothermic reaction is favoured at lower temperature . Hence more of the product (CuCl2) will be formed at a lower temperature of 100C [B]
48.  The rate of reactions involving solid reactants like zinc increases when the solid is in powdered form because this provides a larger surface area for reaction [A]
49.  Zn + H2SO4   ZnSO4 + H2
From the above equation:
1mole of H2SO4 dissolved 65g of Zn
10cm3 of 1.0 M H2SO4 =  ×  = 0.01mole
0.01mole of H2SO4 will dissolve
 ×  = 0.65g of Zinc
Mass of undissolved zinc = (2.00 – 0.65) = 1.35g [A]
50.  Equation for the reaction:
Al(NO3)3 + 3NaOH   3NaNO3 + Al(OH)3
From the above equation:
1 mole of Al(NO3)3 reacted with 3 moles of NaOH. Comparing this with the values provided in the questions:
30cm3 of 0.1M Al(NO3)3 = 0.003mole. This should require (3 × 0.003) = 0.009 mole.
100cm3 of 0.15M NaOH  = 0.015mole. This is greater than 0.009 mole, Thus NaOH is in excess.
Let the actual volume of NaOH that reacted be Vb
Applying the volumetric formula:
        =
Vb =  = 60cm3
Volume of NaOH provided = 100cm3
Volume of excess NaOH = (100 -60) = 40cm3 [C]
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