Mole Concept JEE Advanced Previous Year Questions with Solutions will help JEE aspirants develop better skills to solve different types of problems and prepare effectively for the upcoming entrance exam. We have listed down a set of previous year JEE Advanced questions that have been framed from the topic of the mole concept. Aspirants can therefore go through the questions and get an understanding of the question types, difficulty level and more. Additionally, the detailed solutions that we have provided will further help them to come up with the right approach to the answers.
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JEE Advanced Previous Year Questions on Mole Concept
Question 1. In the chemical reaction between stoichiometric quantities of KMnO4 and KI in a weakly basic solution, what is the number of moles of I2 released for 4 moles of KMnO4 consumed?
Solution: (6)
The chemical reaction of KMnO4 and KI in a weakly basic solution is given as;
Equivalents of KMnO4 = Equivalents of I2
n-factor × Number of moles (n) = n-factor × Number of moles (n)
3 × moles of KMnO4 = 2 × moles of I2
3 × 4 = 2 × moles of I2
Moles of I2 = 6 moles
Question 2. The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by NiCl2.6H2O to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952g of NiCl2.6H2O are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is___. (Atomic weights in g mol–1: H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59)
Solution: (2992)
Therefore, total mass = = 12 × 172 + 4 × 232 = 2992 g
Question 3. Galena (an ore) is partially oxidized by passing air through it at high temperatures. After some time, the passage of air is stopped, but the heating is continued in a closed furnace such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of O2 consumed is ______. (Atomic weights in g mol-1 : O = 16, S = 32, Pb = 207)
Solution: (6.47)
The given reaction will be:
2PbS + 3O2 → 2PbO + 2SO2
2PbO + PbS → 3Pb + SO2
━━━━━━━━━━━━━━
3PbS + 3O2 → 3Pb + 3SO2
So, 3 mol of O2 gives 3 mol of Pb
Thus, in other words, 96 kg of O2 gives 621 kg of Pb.
So, 1 kg of O2 will give 621 / 96 = 6.47kg
Question 4. If the value of the Avogadro number is 6.023 × 1023 and the value of the Boltzmann constant is 1.380 × 10-23 J/K, then the number of significant digits in the calculated value of the universal gas constant is.
Solution: (4)
We have, Universal gas constant,
R = kNA
Where, k = Boltzmann constant and NA = Avogadro’s number
Therefore, R = 1.380 × 10-23 × 6.023 × 1023 J/Kmol
= 8.31174 ≅ 8.312
Since k and NA both have four significant figures, so the value of R is also rounded off up to 4 significant figures.
When the number is rounded off, the number of the significant figure is reduced.
So, the number of the significant figure is 4.
Question 5. Which has the maximum number of atoms?
A. 24 g of C (12)
B. 56 g of Fe (56)
C. 27 g of A1 (27)
D. 108 g of Ag (108)
Solution: (A)
Number of atoms = Number of moles
Avogadro’s number (NA)
Number of atoms in 24 g C = 24 /12 × NA = 2NA
Number of atoms in 56 g of Fe = 56 / 56 NA = NA
Number of atoms in 27 g of Al = 27 / 27 NA= NA
Number of atoms in 108 g of Ag = 108 / 108 NA = NA
Hence, 24 g of carbon has the maximum number of atoms.
Question 6. How many moles of electrons weigh one kilogram?
A. 6.022 x 1023
B. 1 / 9.108 x 1031
C. 6.022 / 9.108 x 1054
D. 1 / 9.108 × 6.022 x 108
Solution: (D)
Mass of one electron = 9.108 × 10-31 kg.
Mass of one mole of electrons =9.108 × 10-31 × 6.023 × 1023 = 9.108 × 6.023 × 10-8 kg.
The number of moles of electrons that weigh one kilogram
= 1 / 9.108 × 6.023 × 10-8 = 1 / 9.108 × 6.023 × 108
Hence, the correct option is D
Question 7. Calculate the molarity of water if its density is 1000 kg/m3.
Solution: (55.56)
Given, the density of the water = 1000 kg/m3
We know, 1 litre of water = 1 kg.
Since the density is 1000 kg/m3
Thus, 1000 ml of water = 1 kg.
(Since, 1 liter = 1000 ml)
Now, Mass of Water = 1 kg.
= 1000 g.
The molecular mass of the water molecules = 18 grams.
Using the formula,
No.of moles = mass / molar mass
= 1000 / 18
= 55.56 moles
For finding Molarity, we use the given formula,
Molarity = Number of moles / volumeinml × 1000
= 55.56 / 1000 × 1000
= 55.56 M
Question 8. Mixture X = 0.02 mole of [Co(NH3)5 SO4] Br and 0.02 mole of [Co(NH3)rBr]SO4 was prepared in 2 L of solution.
1 L of the mixture if X + excess AgNO3 → Y
1 L of mixture X+ excess BaCl2 → Z
Number of moles of y and Z are:
A. 0.01, 0.01
B. 0.02, 0.01
C. 0.01, 0.02
D. 0.02, 0.02
Solution: (A)
In 1 L solution, there will be 0.01 mole of each [Co(NH3)5 SO4] Br and [Co(NH3)rBr]SO4.
Addition of an excess of AgNO3 will give 0.01 mole of AgBr.
Addition of an excess of BaCI2 will give 0.01 mole of BaSO4.
Question 9. An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is
A. 40 mL
B. 20 mL
C. 10 mL
D. 4 ml
Solution: (A)
n-factor for H2C2O4.2H2O = 2
Normality = weight × 2 × 1000 / molecular weight × 250 = 6.3 × 2000 / 126 × 250 = 0.4
equating no. of equivalents:
N1V1 = N2V2
0.4 × 10 = V2 × 0.1
Volume (V2) = 40mL.
Hence, the correct option is A
Question 10. The normality of 0.3 M phosphorous acid H3Po3 is:
A. 0.1
B. 0.9
C. 0.3
D. 0.6
Solution: (D)
Normality = molarity × basicity (for acids)
Therefore, Normality = 0.3 × 2 = 0.6
∵ The basicity of H3PO3 is 2 because one of the H is not replaceable in H3PO3. The H atoms which are linked to oxygen are replaceable, but the H atom linked directly to central atom P is non-replaceable.
Also Read:-
JEE Main Mole Concept Previous Year Questions with Solutions |
Mole Concept Basics |
Mole Concept, Molar Mass and Percentage Composition |
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