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Test: Mole concept - JEE MCQ


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15 Questions MCQ Test - Test: Mole concept

Test: Mole concept for JEE 2024 is part of JEE preparation. The Test: Mole concept questions and answers have been prepared according to the JEE exam syllabus.The Test: Mole concept MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Mole concept below.
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Test: Mole concept - Question 1

1.44 g of titanium (At. mass = 48) reacted with excess of O2 and produce of non stoichiometric compound . The value of is:

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Test: Mole concept - Question 2

The mass of produced by the reaction of of and of is . What is the per cent yield?

Detailed Solution for Test: Mole concept - Question 2


Amount of formed by
Amount of formed by
will be limiting and actual amount of product is

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Test: Mole concept - Question 3

The impure of is dissolved in water and then treated with excess of silver nitrate solution. The mass of precipitate of silver chloride is found to be . The purity of solution would be:

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The reaction that takes place is

∴143.5 g of is produced from
∴ 14 g of will be produced from

This is the amount of in common salt:
purity

Test: Mole concept - Question 4
How many moles of ions are in 20 ml of 0.4 M ?
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Number of moles of
Test: Mole concept - Question 5

What is the empirical formula of vanadium oxide, if 2.74 g of the metal oxide contains 1.53 g of metal?

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Mass of oxide Mass of metal Mass of oxygen

Moles of
Moles of

Test: Mole concept - Question 6

of ferrous ammonium sulphate crystals are dissolved in of water. of this solution requires of potassium permaganate during titration for complete oxidation. The weight of present in one litre of the solution of

Detailed Solution for Test: Mole concept - Question 6

Normality of ferrous amm. sulphate
Eq. wt of FAS is 392


g ev. of
ev. of .

Test: Mole concept - Question 7

Specific volume of cylindrical virus particle is . whose radius and length are respectively.
If , find molecular weight of virus

Detailed Solution for Test: Mole concept - Question 7

Specific volume (volume of ) of cylindrical virus particle
Radius of virus
Length of virus
Volume of virus


Wt. of one virus particle
∴ Mol. wt. of virus Wt. of particle

Test: Mole concept - Question 8

10 moles and 15 moles were allowed to react over a suitable catalyst. 8 moles of were formed. The remaining moles of and respectively are -

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Hence, remaining, SO2 = 10 − 8 = 2 moles,
O2 = 15 − 4 = 11 moles

Test: Mole concept - Question 9

On reduction with hydrogen of an oxide of metal leaves of metallic residue. If the atomic mass of metal is 64 , the formula of metal oxide is

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Mass of oxygen which gets displaced from metal oxide
Now, of oxygen combines with metal
of oxygen combines with metal
GEW of metal
Valency of metal
Hence, formula of oxide is .

Test: Mole concept - Question 10

Vapour density of a metal chloride is 83. If equivalent weight of the metal is 6, its atomic weight will be ____.

Detailed Solution for Test: Mole concept - Question 10

The atomic weight of metal is equal to the product of the equivalent weight of metal and valency. Atomic weight of metal = n × equivalent weight of metal
=4 × 6 = 24
Hence, the correct option (b).

Test: Mole concept - Question 11
The amount of formed upon mixing of solution with 50 of solution will be:
(Given that molecular weight of and
Detailed Solution for Test: Mole concept - Question 11
of solution .
of solution .
The reaction is as follows:

Since, is the limiting reagent, only moles of will form.
Hence, of is formed.
Test: Mole concept - Question 12

9 moles of "D" and 14 moles of E are allowed to react in a closed vessel according to given reactions. Calculate number of moles of B formed in the end of reaction, if 4 moles of G are present in reaction vessel. (Percentage yield of reaction is mentioned in the reaction)
Step-1
Step-2

Detailed Solution for Test: Mole concept - Question 12


9 mole 14 mole mole

Limiting Reagent is
∴ Moles of formed

Test: Mole concept - Question 13

A mixture of and contain mass per cent of nitrogen. What is the mass ratio of the two components in the mixture?

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Let wt. of and are and y gram respectively

Test: Mole concept - Question 14

of on heating gave of and of . This is in accordance with

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mass of reactant = mass of products = 20 g Hence the law of conservation of mass is obeyed.

Test: Mole concept - Question 15

A gaseous mixture contains oxygen and nitrogen in the ratio of 1: 4 by weight. Therefore, the ratio of their number of molecules is

Detailed Solution for Test: Mole concept - Question 15

Let the weight of oxygen
The weight of nitrogen
Number of molecules of oxygen
Number of molecules of nitrogen

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