Chemistry, 1990, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1990

Title :

Chemistry

Exam :

JAMB Exam

Paper 1 | Objectives

31 - 40 of 49 Questions

#	Question	Ans
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	D
32.	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? A. 20 B. 25 C. 50 D. 75	B
33.	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 i A. 50.2 B. 47.0 C. 4.70 D. 0.47 Show Content Detailed Solution 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	C
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	A
35.	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) A. 0.13 B. 0.25 C. 1.25 D. 2.00 Show Content Detailed Solution 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	C
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	B
37.	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 Show Content Detailed Solution 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 10<	A
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 Show Content Detailed Solution M_a = (M_bV_b)/(V_a) = (0.1 m χ 20 cm³)/(25cm) = 0.08 m M = (concentration in g/dm³)/(m) = (8)/(0.08) = 100 g	C
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 Show Content Detailed Solution PH = -log (H⁺) (H⁺) = 10^-4.348 = 4.0 χ 10^-5	A
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	A

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	D
32.	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? A. 20 B. 25 C. 50 D. 75	B
33.	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 i A. 50.2 B. 47.0 C. 4.70 D. 0.47 Show Content Detailed Solution 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	C
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	A
35.	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) A. 0.13 B. 0.25 C. 1.25 D. 2.00 Show Content Detailed Solution 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	C

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	B
37.	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 Show Content Detailed Solution 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 10<	A
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 Show Content Detailed Solution M_a = (M_bV_b)/(V_a) = (0.1 m χ 20 cm³)/(25cm) = 0.08 m M = (concentration in g/dm³)/(m) = (8)/(0.08) = 100 g	C
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 Show Content Detailed Solution PH = -log (H⁺) (H⁺) = 10^-4.348 = 4.0 χ 10^-5	A
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	A