
Mixed MCQ chemistry -2
Quiz
•
Chemistry
•
12th Grade
•
Practice Problem
•
Hard
Geetika Pant
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16 questions
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1.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Primary amine with Hinsberg's reagent forms
N-alkylbenzenesulphonamide soluble in KOH solution.
N-alkylbenzenesulphonamide insoluble in KOH solution.
N,N-dialkylbenzenesulphonamide soluble in KOH solution.
N,N-dialkylbenzenesulphonamide insoluble in KOH solution.
Answer explanation
2.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Which of the following molecules has highest dipole moment?
CH3CI
CH2Cl2
CHCl3
CCl4
Answer explanation
3.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Match the column I with column II and mark the appropriate choice.
(A) → (i), (B) → (ii), (C) → (iii), (D) → (iv)
(A) → (i), (B) → (iii), (C) → (iv), (D) → (i)
(A) → (iv), (B) → (i), (C) → (ii), (D) → (iii)
(A) → (iii), (B) → (iv), (C) → (ii), (D) → (i)
4.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Arrange the following compounds in order of increasing boiling points.
1-Bromopropane, Isopropyl bromide, 1-Bromobutane
1-Bromobutane < 1-Bromopropane < Isopropyl bromide
1-Bromopropane < 1-Bromobutane < Isopropyl bromide
1-Bromopropane < Isopropyl bromide < 1-Bromobutane
Isopropyl bromide < 1-Bromopropane < 1-Bromobutane
Answer explanation
The boiling point of 1-bromobutane is higher than that of 1-bromopropane because as the size of alkyl group increases boiling point increases. The boiling point of isopropyl bromide is lower than that of 1-bromopropane because as the branching increases the boiling point decreases. The boiling points of the given compounds in the increasing order :
Isopropyl bromide < 1-Bromopropane < 1-Bromobutane
5.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
The rate constant for a first order reaction is 7.0 x 10-4 s-1 If initial concentration of reactant is 0.080 M, is the half life of reaction?
990 s
79.2 s
12375 s
10.10 x 10-4 s
Answer explanation
6.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Which is the product of the following reaction?
Answer explanation
7.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
A solution containing 2.675 g of CoCl3.6NH, (molar mass = 267.5 g mol-1) is passed through a cation
exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.78 g of AgCl
(molar mass = 143.5 g mol-1). The formula of the complex is (At. mass of Ag = 108 u)
[CoCI(NH3)5]Cl2
[Co(NH3)6)Cl3
[CoCl2(NH3)4]CI
[CoCl3(NH3)3]
Answer explanation
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