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Test: Laws of chemical combination - JEE MCQ


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15 Questions MCQ Test Chemistry for JEE Main & Advanced - Test: Laws of chemical combination

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

The percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (atomic weight =78.4). Then minimum molecular weight of peroxidase anhydrous enzyme is

Detailed Solution for Test: Laws of chemical combination - Question 1

by weight means if Mol. wt. is 100 then mass of Se is . If at least one atom of Se is present in the molecule then
M.

Test: Laws of chemical combination - Question 2

The number of water molecules present in a drop of water (volume 0.0018 mL) density = 1g mL−1 at room temperature is

Detailed Solution for Test: Laws of chemical combination - Question 2

As we know that,
Density
mass density volume
Density of water
Volume of water (Given)
Mass of water
Molar mass of water
Number of moles in of water
As we know that,
Number of molecules of water in 1 mole molecules
Number of molecules of water in mole
Hence, the number of water molecules present in a drop of water (volume ) at room temperature are .

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Test: Laws of chemical combination - Question 3

A mixture of and gas "Y" mol. mass 80 in the mole ratio a has a mean molecular mass 40. What would be mean molecular mass, if the gases are mixed in the ratio a under, identical conditions? (Assume that gases are non-reacting):

Detailed Solution for Test: Laws of chemical combination - Question 3

Let mole fraction of is
or

When ratio is changed

Test: Laws of chemical combination - Question 4

If molecules are removed from of , then the number of moles of left are

Detailed Solution for Test: Laws of chemical combination - Question 4

Moles of in of

Moles of removes
Moles of left

Test: Laws of chemical combination - Question 5

Arrange the following in the order of increasing mass (atomic mass: O = 16, Cu = 63, N = 14)
I. one atom of oxygen
II. one atom of nitrogen
III. mole of oxygen
IV. mole of copper

Detailed Solution for Test: Laws of chemical combination - Question 5

Mass of atoms of oxygen Mass of one atom of oxygen

Mass of atoms of nitrogen
Mass of one atom of nitrogen

Mass of 1 mole of oxygen Mass of mole of oxygen Mass of 1 mole of copper Mass of mole of copper
So, the order of increasing mass is

Test: Laws of chemical combination - Question 6

Complete combustion of of compound gives of and of . The lowest molecular mass can have:

Detailed Solution for Test: Laws of chemical combination - Question 6

Moles of
mass of
Mole of Moles of
mass of
Compound does not contains oxygen
So EF
Lowest M.M.

Test: Laws of chemical combination - Question 7

When burnt in air, 14.0 g mixture of carbon and sulphur gives a mixture of CO2 and SO2 in the volume ratio of 2:1, volume being measured at the same conditions of temperature and pressure moles of carbon in the mixture is

Detailed Solution for Test: Laws of chemical combination - Question 7

Let weight of be , then will be

∴ x = 6 g; Moles of C = 6/12 = 0.5

Test: Laws of chemical combination - Question 8

1.12 mL of a gas is produced at S.T.P. by the action of 4.12mg of alcohol ROH with methyl magnesium Iodide. The molecular mass of alcohol is

Detailed Solution for Test: Laws of chemical combination - Question 8

Let the alcohol be
ROH and x its molecular weight

Test: Laws of chemical combination - Question 9

1 mole of mixture of CO and CO2 requires exactly 28 gKOH in solution for complete conversion of all the CO2 into K2CO3. How much amount more of KOH will be required for conversion into K2CO3 if one mole of mixture is completely oxidized to CO2.

Detailed Solution for Test: Laws of chemical combination - Question 9


It corresponds to
0.25 mol of CO2
Hence mol of CO = 1 − 0.25 = 0.75 ≡ mole of
CO2 formed Mol of KOH requred = 2 × 0.75 = 1.5
= 1.5 × 56 = 84 g

Test: Laws of chemical combination - Question 10

1 mL of water has 25 drops. Let N0 be the Avogadro number. What is the number of molecules present in 1 drop of water ? (Density of water = 1 g/mL)

Detailed Solution for Test: Laws of chemical combination - Question 10

Volume of one drop
Mass of


Number of moles of
Number of Molecule

Test: Laws of chemical combination - Question 11

A 25.0 mm×40.0 mm piece of gold foil is 0.25 mm thick. The density of gold is 19.32 g/cm3. How many gold atoms are in the sheet? (Atomic weight : Au = 197.0)

Detailed Solution for Test: Laws of chemical combination - Question 11

Volume of gold foil

Mass of gold foil

No. of gold atoms

Test: Laws of chemical combination - Question 12

of and of are made to react completely to yield a mixture of and Calculate moles of ICl and formed

Detailed Solution for Test: Laws of chemical combination - Question 12

Test: Laws of chemical combination - Question 13

The largest number of molecules is in

Detailed Solution for Test: Laws of chemical combination - Question 13

Number of molecules in of water

Number of molecules in of

Number of molecules in of

Number of molecules in of

Hence, in these of water has largest number of molecules.

Test: Laws of chemical combination - Question 14

of oxygen contains number of atoms equal to that in

Detailed Solution for Test: Laws of chemical combination - Question 14

O atoms in 2g of oxygen

 of sulphur also contains atom = 0.125 NA

Test: Laws of chemical combination - Question 15
If we consider that , in place of , mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of the substance will
Detailed Solution for Test: Laws of chemical combination - Question 15
Let's call the new unit amu'. Taking this to be the standard, the mass of 1 atom would be times the old amu - hence, ', but hold on, the mass of 1 mole of carbon will be the mass of carbon in amu'* mass of amu'* avagadro number (mass of amu in the old mass itself!
So mass remains unchanged.
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