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Test: Concentration, Empirical & Molecular Formulae - NEET MCQ


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17 Questions MCQ Test Chemistry Class 11 - Test: Concentration, Empirical & Molecular Formulae

Test: Concentration, Empirical & Molecular Formulae for NEET 2024 is part of Chemistry Class 11 preparation. The Test: Concentration, Empirical & Molecular Formulae questions and answers have been prepared according to the NEET exam syllabus.The Test: Concentration, Empirical & Molecular Formulae MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Concentration, Empirical & Molecular Formulae below.
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Test: Concentration, Empirical & Molecular Formulae - Question 1

For the reaction 2x + 3y + 4z → 5w

Initially if 1 mole of x, 3 mole of y and 4 mole of z is taken. If 1.25 mole of w is obtained then % yield of this reaction is

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 1

1 moles of x will give = 5/2 = 2.5 mol

Test: Concentration, Empirical & Molecular Formulae - Question 2

Molarity of NaOH in a solution prepared by dissolving 4 g of NaOH in enough water to form 250 ml of solution is:​

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 2

Weight of NaOH = 4g

Molecular weight of NaOH = 40G

Volume of solution is 250ml

Molarity = ?
M = WMW × 1000 V in ml

M = 440 × 1000250 = 0.4M

Test: Concentration, Empirical & Molecular Formulae - Question 3

125 ml of 8% w/w NaOH solution (sp. gravity 1) is added to 125 ml of 10% w/v HCl solution. The nature of resultant solution would be ____

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 3

= 0.25 mole

Test: Concentration, Empirical & Molecular Formulae - Question 4

Ratio  of masses of H2SO4  and Al2 (SO4)3 is grams each containing 32 grams of S is _____

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 4

Test: Concentration, Empirical & Molecular Formulae - Question 5

18 g of glucose (C6H12O6) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is:​

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 5

18 g of glucose (C6H12O6) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is 0.1 M.

Test: Concentration, Empirical & Molecular Formulae - Question 6

For the reaction 2A + 3B + 5C → 3D

Initially if 2 mole of A, 4 mole of B and 6 mole of C is taken, With 25% yield, moles of D which can be produced are _____________.

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 6

Limiting reactant is A
Ideally with 2 moles of A, D formed = 3 moles But yield = 25%
So, moles of D formed = 3 × 0.25 = 0.75 mol

Test: Concentration, Empirical & Molecular Formulae - Question 7

Two elements X (atomic mass = 75) and Y (atomic mass = 16) combine to give a compound having 75.8% of X. The formula of the compound is:

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 7

So empirical formula of compound = X₂Y₃

For a molecular formula you have to provide molecular mass of the compound.

Test: Concentration, Empirical & Molecular Formulae - Question 8

Equal volumes of 10% (v/v) of HCl is mixed with 10% (v/v) NaOH solution. If density of pure NaOH is 1.5 times that of pure HCl then the resultant solution be :

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 8

Both have equal volume = V

NaOH mole > HCl mole
Basic Solution

Test: Concentration, Empirical & Molecular Formulae - Question 9

Similar to the % labelling of oleum, a mixture of H3PO4 and P4O10 is labelled as (100 + x) % where x is the maximum mass of water which can react with P4O10 present in 100 gm mixture of H3PO4 and P4O10. If such a mixture is labelled as 127% Mass of P4O10 is 100 gm of mixture, is

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 9

P4O10 + 6 H2O → 4H3 PO4
284 gm 108gm 392 gm 108 gm water reacts with P4O10 = 284 gm
27 gm water will react with P4O10 = 71 gm

Test: Concentration, Empirical & Molecular Formulae - Question 10

C6H5OH(g) + O2(g) → CO2(g) + H2O(l)

The magnitude of volume change if 30 ml of C6H5OH (g) is burnt with an excess amount of oxygen, is

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 10

C6H5OH(g)+7O2(g) → 6CO2(g)+3H2O(l)
30 ml
6 x 30= 180 ml of CO2 is produced
Volume used initially  = 30 + 210 = 240
(for C6H5OH)(For O2)
change in volume = 240 - 180 = 60 ml

Test: Concentration, Empirical & Molecular Formulae - Question 11

How many moles of NaCl should be dissolved in 100 mL of water to get 0.2 M solution?​

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 11

Molarity = no of moles /volume( in litre) 0.2 =no of moles / 0.1 No of moles =0.2 × 0.1 = 0.02

Test: Concentration, Empirical & Molecular Formulae - Question 12

If 50 gm oleum sample rated as 118% is mixed with 18 gm water, then the correct option is

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 12

118 % ⇒ 100 g sample uses 18 g water
⇒ 50 g sample need 9 g water
(50 g + 18 g water)
⇒ 59 g H2SO4 + 9 g water

Test: Concentration, Empirical & Molecular Formulae - Question 13

The minimum mass of mixture of A2 and B4 required to produce at least 1 kg of each product is:
(Given At. mass of ‘A’ = 10; At mass of ‘B’ = 120) 5A2 + 2B4 → 2AB2  + 4A2B

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 13


Here limiting product is AB2 = 500 gm

So, A2 needed = 10 × 20 = 200 gm
B2 needed = 480 × 4 = 1920 gm
Total mass of mixture = 2120 gm

Test: Concentration, Empirical & Molecular Formulae - Question 14

The % by volume of C4H10 in a gaseous mixture of C4H10, CH4 and CO is 40. When 200 ml of the mixture is burnt in excess of O2. Find volume (in ml) of CO2 produced.

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 14

C4H10 = 80 ml
CH4 = xml  CO = y ml
x + y = 120 ml

total CO2 volume
= 320 + x + y ml
= 320 + 120
= 440 ml

Test: Concentration, Empirical & Molecular Formulae - Question 15

74 gm of sample on complete combustion gives 132 gm CO2 and 54 gm of H2O. The molecular formula of the compound may be

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 15



⇒ C atoms = 3 mole
H atoms = 6 mole

Test: Concentration, Empirical & Molecular Formulae - Question 16

Density of a gas relative to air is 1.17. Find the mol. mass of the gas. [Mair = 29 g/mol]

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 16



M gas = 29 × 1.17 = 33.9

Test: Concentration, Empirical & Molecular Formulae - Question 17

Number of moles of hydroxide ions in 0.3L of 0.005M solution of Ba (OH)2 is:

Detailed Solution for Test: Concentration, Empirical & Molecular Formulae - Question 17

0.3 L = 0.3 L of 0.005 M Ba(OH)2
0.3 x 0.005 = 0.0015 moles of Ba(OH)2
It will dissociate into 0.0015 moles of Ba and 2 x 0.0015 = 0.003 moles of OH ions.

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