1999 - JAMB Chemistry Past Questions and Answers - page 1
1
200cm3 each of 0.1M solutions of lead (II) trioxonitrate (V) and hydrochloric acid were mixed. Assuming that lead (II) chloride is completely insoluble, calculate the mass of lead (II) chloride that will be precipitated.
[Pb = 207, Cl= 35.5, N = 14, O = 16]
[Pb = 207, Cl= 35.5, N = 14, O = 16]
A
2.78g
B
5.56g
C
8.34g
D
11.12g
correct option: b
The balanced equation of the reaction is:
Pb(NO3)2(aq) + 2HCl(aq) ---> PbCl2(s) + 2HNO3(aq)
From the balanced equation,
1 mole of Pb(NO3)2 gave 1 mole of PbCl2
using, molar concentration = no of moles/volume in dm3
therefore, no of moles = molar concentration x volume in dm3
where, molar concentration of Pb(NO3)2 = 0.1M
volume of Pb(NO3)2 in dm3 = 0.2 dm3
therefore, no of moles of Pb(NO3)2 = 0.1M x 0.2 dm3 = 0.02 mol
but from the balanced equation of the reaction, 1mol of Pb(NO3)2 gave 1mol of PbCl2,
therefore 0.02mol of Pb(NO3)2 will give 0.02mol of PbCl2.
But using, no of moles = mass/molar mass,
therefore mass = no of moles x molar mass
where, no of moles of PbCl2 = 0.02mol
molar mass of PbCl2 = 207 + (2 * 35.5) = 278 g/mol
therefore, mass of PbCl2 = 0.02mol * 278g/mol
= 5.56g
therefore, mass of PbCl2 that precipitated out is 5.56g.
Working and Answer was provided by
Users' Answers & CommentsPb(NO3)2(aq) + 2HCl(aq) ---> PbCl2(s) + 2HNO3(aq)
From the balanced equation,
1 mole of Pb(NO3)2 gave 1 mole of PbCl2
using, molar concentration = no of moles/volume in dm3
therefore, no of moles = molar concentration x volume in dm3
where, molar concentration of Pb(NO3)2 = 0.1M
volume of Pb(NO3)2 in dm3 = 0.2 dm3
therefore, no of moles of Pb(NO3)2 = 0.1M x 0.2 dm3 = 0.02 mol
but from the balanced equation of the reaction, 1mol of Pb(NO3)2 gave 1mol of PbCl2,
therefore 0.02mol of Pb(NO3)2 will give 0.02mol of PbCl2.
But using, no of moles = mass/molar mass,
therefore mass = no of moles x molar mass
where, no of moles of PbCl2 = 0.02mol
molar mass of PbCl2 = 207 + (2 * 35.5) = 278 g/mol
therefore, mass of PbCl2 = 0.02mol * 278g/mol
= 5.56g
therefore, mass of PbCl2 that precipitated out is 5.56g.
Working and Answer was provided by
2
56.00cm3 of a gas at S.T.P. weighed 0.11g. What is the vapour density of the gas?
[Molar volume of a gas at S.T.P = 22.4dm3]
[Molar volume of a gas at S.T.P = 22.4dm3]
A
11.00
B
22.00
C
33.00
D
44.00
correct option: b
56.00cm3 of a gas at S.T.P weighed 0.11g
Molar volume of the gas = 22.4dm3 = 22400cm3
molar mass = (22400 x 0.11)/56
molar mass = 2464/56 = 44
Vapour density = Molar mass/2 = 44/2 = 22
Users' Answers & CommentsMolar volume of the gas = 22.4dm3 = 22400cm3
molar mass = (22400 x 0.11)/56
molar mass = 2464/56 = 44
Vapour density = Molar mass/2 = 44/2 = 22
3
Which of the following gases will diffuse fastest when passed through a porous plug?
[H = 1, C = 12, N = 14, O = 16]
[H = 1, C = 12, N = 14, O = 16]
A
Propane
B
Oxygen
C
Methane
D
Ammonia
correct option: c
Users' Answers & Comments4
Which is the temperature of a given mass of a gas initially at 0°C and 9 atm, if the pressure is reduced to 3 atm at constant volume?
A
91K
B
182K
C
273K
D
819K
correct option: a
From P1/T1 = P2/T2 = 9/273 = 3/T2
T2 x 9 = 273 x 3
T2 = (273 x 3)/9
T2 = 91K
Users' Answers & CommentsT2 x 9 = 273 x 3
T2 = (273 x 3)/9
T2 = 91K
5
Which of the following will have its mass increased when heated in air?
A
Helium
B
Magnesium
C
Copper pyrites
D
Glass
correct option: b
Users' Answers & Comments6
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
From the equation above, the mass of magnesium required to react with 250cm3 of 0.5M HCl is
[Mg = 24]
From the equation above, the mass of magnesium required to react with 250cm3 of 0.5M HCl is
[Mg = 24]
A
0.3g
B
1.5g
C
2.4g
D
3.0g
correct option: b
Users' Answers & Comments7
A gaseous metallic chloride MClx consists of 20.22% of M by mass. The formula of the chloride is
[M = 27, Cl = 35.5]
[M = 27, Cl = 35.5]
A
MCl
B
MCl2
C
MCl3
D
M2Cl6
correct option: c
Percentage by mass of M = 20.22%
Percentage by mass of Cl = ?
100 - 20.22 = 79.78%
Mass given ration of M
M = 2022/27 = 0.75
Chlorine = 79.78/35.5 = 2.24
Divide by the smallest
M = 0.75/0.75 = 1
Cl = 2.24/0.75 = 3
∴ The formula = MCl3
Users' Answers & CommentsPercentage by mass of Cl = ?
100 - 20.22 = 79.78%
Mass given ration of M
M = 2022/27 = 0.75
Chlorine = 79.78/35.5 = 2.24
Divide by the smallest
M = 0.75/0.75 = 1
Cl = 2.24/0.75 = 3
∴ The formula = MCl3
8
In which of the following are water molecules in the most disorderly arrangement?
A
Ice at -10°C
B
Ice at 0°C
C
Water at 100°C
D
Steam at 100°C
correct option: d
Users' Answers & Comments9
In order to remove one electron from a 3S orbital of gaseous sodium atom, about 496KJMol-1 of energy is required. This energy is referred to as
A
electron affinity
B
ionization energy
C
activation evergy
D
electronegativity
correct option: b
Users' Answers & Comments10
Nitrogen obtained from the liquefaction of air has a higher density than that obtained from nitrogen containing compounds because the former contains
A
water vapour
B
oxygen
C
carbon (IV) oxide
D
rare gases
correct option: d
Users' Answers & Comments