Year :
1982
Title :
Chemistry
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 49 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
When a little brine was added to a metal nitrate a white precipitate was formed which first turned violet and then black on exposure to ai. The metal nitrate is therefore? A. sodium nitrate B. mercuric nitrate C. lithium nitrate D. silver nitrate E. potassium nitrate |
D |
| 22. |
If atom X with six electrons in its outermost shell combines with atom Y with one electron in its outermost shell to form an ionic compound, then? A. six atoms of X will combine with one of Y B. two atoms of X will combine with six of Y C. one atom of X will combine with two of Y D. two atoms of X will combine with one of Y E. one atom of X will combine with six of Y Detailed Solution- When an element in group 1 takes part in a reaction, its atoms lose their outer electron and form positively charged ions , called cations .- Group 6 elements, have 6 valence electrons and will tend to take 2 electrons and have a valency of -2. - Hence; one atom of element X will require two atoms of Y - To form a compound XY\(_2\) |
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| 23. |
The nitrogen gas obtained by burning an excess of phosphorus in air has a density higher than the nitrogen from nitrogen compounds because the nitrogen from air is contaminated with? A. some unburnt phosporus B. P2O5 vapour C. some oxygen D. rare gases E. carbondioxide |
D |
| 24. |
When air is passed through alkaline pyrogallol and then over quicklime, the only gases left are? A. nitrogen and carbondoxide B. the rare gases C. nitrogen and oxygen D. nitrogen and the rare gases E. nitrogen, carbonioxide and the rare gases |
D |
| 25. |
9.8g of solid potassium chlorate (KCIO3) were added to 40 cm3 of water and heated to dissolve all the solid. As he solution cooled, crystals of potassium chlorate started forming at 60°C. The solubility of potassium chlorate at 66°C is therefore? A. 24.5 mole per dm3 B. 0.2 mole per dm3 C. 10.0 mole per dm3 D. 20.0 mole per dm3 E. 2.0 mole per dm3 Detailed Solution9.8 gms KCIO3 dissolves in 40 cm3 of H.In 100 cc we have (9.8)/(40) x (100)/(1) = 24.5 gm KCIO3 In 1000 cc we have 10 x 24.5 or 245 gms of KCIO3. But mol wt of KCIO3 is 122.5 Solubility of KCIO3 is (245)/(122.5) = 2 moles per dm3 (E) |
|
| 26. |
100 cm3 of oxygen is made to react with 50 cm 3 of hydrogen, and the whole reaction mixture was then passed through anhydrous calcium chloride. What is the volume of gas left? A. 5 cm3 B. 25 cm3 C. 50 cm3 D. 75 cm3 E. 100 cm3 Detailed Solution2H2 + O2 → 2H2O2V3 1V3 2V 50cm 100 — 50cm 25cm3 50cm3 Volume of gas left = (100 - 25)cm3 of O2 = 75 cm3 of O2 |
|
| 27. |
A mixture contains 20cm3 of hydrogen, 35cm3 of oxygen, 15cm3 of carbon monoxide and 10cm3 of nitrogen at S.T.P. Which of the following gives the mole fraction of hydrogen in this mixture? A. 0.02 B. 0.16 C. 0.20 D. 0.25 E. 20 Detailed SolutionH2 = 20cmO2 = 35cm N2 = 10cm CO = 15cm Total vol = 80cm Mole fraction of H2 = (20)/(80) = 0.25 |
|
| 28. |
0.07g of a hydride of carbon occupies 56.0cm3 at S.T.P. when vaporized and contains 14.29% by mass of hydrocarbon is(C = 12, H = 1) A. CH4 B. C2H2 C. C2H4 D. C2H6 E. C3H8 Detailed Solution% of H2 = 14.29% of C = 85.71 Divided by At wts For H = (14.29)/(1) = 14.29 For C = (85.71)/(12) = 7.14 Dividing by smallest number we have the following: For H (14.29)/(7.14) = 2 For C = (7.14)/(7.14) = 1 Emperical formula of the comp = CH2 wt of hydride taht will occupy 22400 cm is 22400 x 0.7 56 = 28 = mole wt of hydride X(CH2) = 28 X(12 + 2) = 28 i.e 14x = 28x = 2 The formula of hydrocarbon is therefore C2H4 |
|
| 29. |
The pressure on 100 cm3 of oxygen gas at 35°C is 750mm of Hg.What would be the volume of the gas if the pressure is increased to 1000mm of Hg without changing the temperature? A. 133.3 cm3 B. 85 cm3 C. 75 cm3 D. 65 cm3 E. 58.3 cm3 Detailed SolutionP1 = 750MHg V1 = 100cm3 l2 = 308kP2 = 100 mmHg V2 = ? But (P1V1)/(T1) = ((P2V2)/(T2) i.e (750 x 100)/(308) = (1000 x )/(V2)/(308) V2 = (750 x 1000 x 308)/(308 x 1000) = 75 cm3 |
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| 30. |
Which of the following bonds exist in crystalline ammonium chloride (NH4Cl)? A. Ionic and covalent B. Ionic and co-ordinate C. Ionic covalent and co-ordinate D. Covalent , co-ordinate and metalllic E. Ionic, covalent and metallic |
C |
| 21. |
When a little brine was added to a metal nitrate a white precipitate was formed which first turned violet and then black on exposure to ai. The metal nitrate is therefore? A. sodium nitrate B. mercuric nitrate C. lithium nitrate D. silver nitrate E. potassium nitrate |
D |
| 22. |
If atom X with six electrons in its outermost shell combines with atom Y with one electron in its outermost shell to form an ionic compound, then? A. six atoms of X will combine with one of Y B. two atoms of X will combine with six of Y C. one atom of X will combine with two of Y D. two atoms of X will combine with one of Y E. one atom of X will combine with six of Y Detailed Solution- When an element in group 1 takes part in a reaction, its atoms lose their outer electron and form positively charged ions , called cations .- Group 6 elements, have 6 valence electrons and will tend to take 2 electrons and have a valency of -2. - Hence; one atom of element X will require two atoms of Y - To form a compound XY\(_2\) |
|
| 23. |
The nitrogen gas obtained by burning an excess of phosphorus in air has a density higher than the nitrogen from nitrogen compounds because the nitrogen from air is contaminated with? A. some unburnt phosporus B. P2O5 vapour C. some oxygen D. rare gases E. carbondioxide |
D |
| 24. |
When air is passed through alkaline pyrogallol and then over quicklime, the only gases left are? A. nitrogen and carbondoxide B. the rare gases C. nitrogen and oxygen D. nitrogen and the rare gases E. nitrogen, carbonioxide and the rare gases |
D |
| 25. |
9.8g of solid potassium chlorate (KCIO3) were added to 40 cm3 of water and heated to dissolve all the solid. As he solution cooled, crystals of potassium chlorate started forming at 60°C. The solubility of potassium chlorate at 66°C is therefore? A. 24.5 mole per dm3 B. 0.2 mole per dm3 C. 10.0 mole per dm3 D. 20.0 mole per dm3 E. 2.0 mole per dm3 Detailed Solution9.8 gms KCIO3 dissolves in 40 cm3 of H.In 100 cc we have (9.8)/(40) x (100)/(1) = 24.5 gm KCIO3 In 1000 cc we have 10 x 24.5 or 245 gms of KCIO3. But mol wt of KCIO3 is 122.5 Solubility of KCIO3 is (245)/(122.5) = 2 moles per dm3 (E) |
| 26. |
100 cm3 of oxygen is made to react with 50 cm 3 of hydrogen, and the whole reaction mixture was then passed through anhydrous calcium chloride. What is the volume of gas left? A. 5 cm3 B. 25 cm3 C. 50 cm3 D. 75 cm3 E. 100 cm3 Detailed Solution2H2 + O2 → 2H2O2V3 1V3 2V 50cm 100 — 50cm 25cm3 50cm3 Volume of gas left = (100 - 25)cm3 of O2 = 75 cm3 of O2 |
|
| 27. |
A mixture contains 20cm3 of hydrogen, 35cm3 of oxygen, 15cm3 of carbon monoxide and 10cm3 of nitrogen at S.T.P. Which of the following gives the mole fraction of hydrogen in this mixture? A. 0.02 B. 0.16 C. 0.20 D. 0.25 E. 20 Detailed SolutionH2 = 20cmO2 = 35cm N2 = 10cm CO = 15cm Total vol = 80cm Mole fraction of H2 = (20)/(80) = 0.25 |
|
| 28. |
0.07g of a hydride of carbon occupies 56.0cm3 at S.T.P. when vaporized and contains 14.29% by mass of hydrocarbon is(C = 12, H = 1) A. CH4 B. C2H2 C. C2H4 D. C2H6 E. C3H8 Detailed Solution% of H2 = 14.29% of C = 85.71 Divided by At wts For H = (14.29)/(1) = 14.29 For C = (85.71)/(12) = 7.14 Dividing by smallest number we have the following: For H (14.29)/(7.14) = 2 For C = (7.14)/(7.14) = 1 Emperical formula of the comp = CH2 wt of hydride taht will occupy 22400 cm is 22400 x 0.7 56 = 28 = mole wt of hydride X(CH2) = 28 X(12 + 2) = 28 i.e 14x = 28x = 2 The formula of hydrocarbon is therefore C2H4 |
|
| 29. |
The pressure on 100 cm3 of oxygen gas at 35°C is 750mm of Hg.What would be the volume of the gas if the pressure is increased to 1000mm of Hg without changing the temperature? A. 133.3 cm3 B. 85 cm3 C. 75 cm3 D. 65 cm3 E. 58.3 cm3 Detailed SolutionP1 = 750MHg V1 = 100cm3 l2 = 308kP2 = 100 mmHg V2 = ? But (P1V1)/(T1) = ((P2V2)/(T2) i.e (750 x 100)/(308) = (1000 x )/(V2)/(308) V2 = (750 x 1000 x 308)/(308 x 1000) = 75 cm3 |
|
| 30. |
Which of the following bonds exist in crystalline ammonium chloride (NH4Cl)? A. Ionic and covalent B. Ionic and co-ordinate C. Ionic covalent and co-ordinate D. Covalent , co-ordinate and metalllic E. Ionic, covalent and metallic |
C |