1990 - JAMB Chemistry Past Questions and Answers - page 4
31
X(g) → X(g). The type of energy involved in the above transformation is?
A
ionization energy
B
sublimation energy
C
lattice energy
D
electron affinity
correct option: d
Users' Answers & Comments32
Chlorine, consisting of two isotopes of mass numbers 35 and 37, has an atomic mass of 35.5.
The relative abundance of the isotope of mass number 37 is?
The relative abundance of the isotope of mass number 37 is?
A
20
B
25
C
50
D
75
correct option: b
Users' Answers & Comments33
10.0 dm3 of air containing H2S as an impurity was passed through a solution of Pb(NO3)2 until all the H2S had reacted. The precipitate of PbS was found to weigh 5.02 g. According to the equation:
Pb(NO3)2 + H2S → PbS + 2HNO3 the percentage by volume of hydrogen sulphide in the air is?
(Pb = 207, S = 32, GMV at s.t.p = 22.4dm3)
Pb(NO3)2 + H2S → PbS + 2HNO3 the percentage by volume of hydrogen sulphide in the air is?
(Pb = 207, S = 32, GMV at s.t.p = 22.4dm3)
A
50.2
B
47.0
C
4.70
D
0.47
correct option: c
Pb(NO3)2 + H2S → PbS + 2HNO3
34 g H2S → 239 g pbs
5.02g pbs → (34)/(237g) χ 5.02 g = 0.714 g
34g → 22.4 dm3
=0.714 → (22.4)/34g) χ 100 = 4.70
Users' Answers & Comments34 g H2S → 239 g pbs
5.02g pbs → (34)/(237g) χ 5.02 g = 0.714 g
34g → 22.4 dm3
=0.714 → (22.4)/34g) χ 100 = 4.70
34
A blue solid, T, which weighed 5.0g was placed on a table. After 8 hours, the resulting pink solid was found to weigh 5.5g. It can be inferred that substance T?
A
is deliquescent
B
is hydroscopic
C
has some molecules of water of crystallization
D
is efflorescent
correct option: a
Users' Answers & Comments35
The solubility in moles per dm3 of 20g of CuSO4 dissolve in 100 of water at 180°C is?
(Cu = 63.5, S = 32, O = 16)
(Cu = 63.5, S = 32, O = 16)
A
0.13
B
0.25
C
1.25
D
2.00
correct option: c
Solubility in moles per dm3
No of moles = (20)/(159.5) χ 0.125 moles
in 100 g of moles = (0.125)/(100) χ 1000 = 1.25 moles
Users' Answers & CommentsNo of moles = (20)/(159.5) χ 0.125 moles
in 100 g of moles = (0.125)/(100) χ 1000 = 1.25 moles
36
Smoke consist of?
A
solid particles dispersed in liquid
B
solid or liquid particles dispersed in gas
C
gas or liquid particles dispersed inliquid
D
liquid particles dispersed inliquid
correct option: b
Users' Answers & Comments37
Na2C2O4 + CaCl2 → CaC2O4 + 2NaCl. Given a solution of 1.9g of sodium oxalate in 50g of water at room temperature, calculate the minimum volume of 0.1 M calcium oxalate using the above equation?
A
1.40 χ 102 dm3
B
14.0 χ 102 cm3
C
1.40 χ 10-2 dm-2
D
14.0 χ 10-2 cm3
correct option: a
Na2C2O4 + CaCl2 → CaC2O4 + 2NaCl
molar mass of Na2C2O4 = 134
molarity in 1000g H2O = (1.9)/(134) = 0.014 m
but in 502 g mole = (0.014)/(50) χ 1000 = 0.28 m
M1V1 = M2V2
V2 = (M1V1)/(M2) = (0.28 χ 50)/(0.1) = 140
= 1.40 χ 102 dm3
Users' Answers & Commentsmolar mass of Na2C2O4 = 134
molarity in 1000g H2O = (1.9)/(134) = 0.014 m
but in 502 g mole = (0.014)/(50) χ 1000 = 0.28 m
M1V1 = M2V2
V2 = (M1V1)/(M2) = (0.28 χ 50)/(0.1) = 140
= 1.40 χ 102 dm3
38
2.0g of a monobasic acid was made up to 250 cm3 with distilled water. 25.00 cm3 of this solution required 20.00cm3 of 0.1 M NaOH solution for complete neutralization. The molar mass of the acid is?
A
200 g
B
160 g
C
100 g
D
50 g
correct option: c
Ma = (MbVb)/(Va) = (0.1 m χ 20 cm3)/(25cm) = 0.08 m
M = (concentration in g/dm3)/(m) = (8)/(0.08) = 100 g
Users' Answers & CommentsM = (concentration in g/dm3)/(m) = (8)/(0.08) = 100 g
39
What is the concentration of H+ ions in moles per dm3 of a sodium of pH 4.398?
A
4.0 χ 10-5
B
0.4 χ 10-5
C
4.0 χ 10-3
D
0.4 χ 10-3
40
What volume of 11.0 M hydrochloric acid must be dilute to obtain 1 dm3 of 0.05 M acid?
A
0.05 dm3
B
0.10 dm3
C
0.55 dm3
D
11.0 dm3
correct option: a
Users' Answers & Comments