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All questions of Some Basic Concepts of Chemistry for JEE Exam
The oxidation number of carbon in CH2O is (1982 - 1 Mark)- a)– 2
- b)+ 2
- c)0
- d)+ 4
Correct answer is option 'C'. Can you explain this answer?
The oxidation number of carbon in CH2O is (1982 - 1 Mark)
a)
– 2
b)
+ 2
c)
0
d)
+ 4
|
Krishna Iyer answered |
NOTE : The sum of oxidation states of all atoms in compound is zero.
Calculation of O.S. of C in CH2O. x + 2 + (–2) = 0 ⇒ x = 0
Calculation of O.S. of C in CH2O. x + 2 + (–2) = 0 ⇒ x = 0
The largest number of molecules is in (1979)- a)36 g of water
- b)28 g of carbon monoxide
- c)46 g of ethyl alcohol
- d)54 g of nitrogen pentoxide
Correct answer is option 'A'. Can you explain this answer?
The largest number of molecules is in (1979)
a)
36 g of water
b)
28 g of carbon monoxide
c)
46 g of ethyl alcohol
d)
54 g of nitrogen pentoxide
|
Pranav Pillai answered |
(a) 18 g of H2O = 6.02 × 1023 molecules of H2O
∴ 36 g of H2O = 2 × 6.02 × 1023 molecules of H2O
= 12.04 × 1023 molecules of H2O
(b) 28 g of CO = 6.02 × 1023 molecules of CO
(c) 46 g of C2H5OH = 6.02 × 1023 molecules of C2H5OH
(d) 108 g of N2O5 = 6.02 × 1023 molecules of N2O5
∴ 54 g of N2O5 =
× 6.02 × 1023 molecules of N2O5
= 3.01 × 1023 molecules of N2O5
∴ 36 g of water has highest number of molecules.
∴ 36 g of H2O = 2 × 6.02 × 1023 molecules of H2O
= 12.04 × 1023 molecules of H2O
(b) 28 g of CO = 6.02 × 1023 molecules of CO
(c) 46 g of C2H5OH = 6.02 × 1023 molecules of C2H5OH
(d) 108 g of N2O5 = 6.02 × 1023 molecules of N2O5
∴ 54 g of N2O5 =
× 6.02 × 1023 molecules of N2O5 = 3.01 × 1023 molecules of N2O5
∴ 36 g of water has highest number of molecules.
How many moles of electron weigh one kilogram? (2002S) - a)6.023 x1023
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
How many moles of electron weigh one kilogram? (2002S)
a)
6.023 x1023
b)
c)
d)
|
Tejas Verma answered |
TIPS/Formulae : (i) Mass of one electron = 9.108 × 10–31 kg
(ii) 1 mole of electron = 6.023 × 1023 electrons
Weight of 1 mole of electron = Mass of one electron × Avogadro Number = 9.108 × 10–31 × 6.023 × 1023 kg
∴ No. of moles of electrons in 1 kg
(ii) 1 mole of electron = 6.023 × 1023 electrons
Weight of 1 mole of electron = Mass of one electron × Avogadro Number = 9.108 × 10–31 × 6.023 × 1023 kg
∴ No. of moles of electrons in 1 kg
The equivalent weight of MnSO4 is half of its molecular weight when it is converted to : (1988 - 1 Mark)- a)Mn2O3
- b)MnO2
- c)MnO4-
- d)MnO42-
Correct answer is option 'B'. Can you explain this answer?
The equivalent weight of MnSO4 is half of its molecular weight when it is converted to : (1988 - 1 Mark)
a)
Mn2O3
b)
MnO2
c)
MnO4-
d)
MnO42-
|
Geetika Shah answered |
For equivalent weight of MnSO4 to be half of its molecular weight, change in oxidation state must be equal to 2. It is possible only when oxidation state of Mn in product is + 4. Since oxidation state of Mn in MnSO4 is + 2. So, MnO2 is correct answer.
In MnO2, O.S. of Mn = +4
∴ Change in O.S. of Mn = +4 – (+2) = +2
In MnO2, O.S. of Mn = +4
∴ Change in O.S. of Mn = +4 – (+2) = +2
The reaction, 3ClO-(aq) → ClO 3- (aq) + 2Cl-(aq), is an example of (2001S)- a)oxidation reaction
- b)reduction reaction
- c)disproportionation reaction
- d)decomposition reaction
Correct answer is option 'C'. Can you explain this answer?
The reaction, 3ClO-(aq) → ClO 3- (aq) + 2Cl-(aq), is an example of (2001S)
a)
oxidation reaction
b)
reduction reaction
c)
disproportionation reaction
d)
decomposition reaction
|
|
Geetika Shah answered |
TIPS/Formulae : (i) In a disproportionation reaction same elemen t undergoes oxidation as well as reduction during the reaction.
(ii) In decomposition reaction a molecule breaks down to more than one atoms or molecules
(ii) In decomposition reaction a molecule breaks down to more than one atoms or molecules
It is disproportionation reaction because Cl is both oxidised (+ 1 to + 5) and reduced ( + 1 to – 1) during reaction.
The total number of electrons in one molecule of carbon dioxide is (1979)- a)22
- b)44
- c)66
- d)88
Correct answer is option 'A'. Can you explain this answer?
The total number of electrons in one molecule of carbon dioxide is (1979)
a)
22
b)
44
c)
66
d)
88
|
Mohit Patel answered |
No. of e– in C = 6 and in O = 8
∴ Total no. of e– in CO2 = 6 + 8 × 2 = 22
∴ Total no. of e– in CO2 = 6 + 8 × 2 = 22
The normality of 0.3 M phosphorous acid (H3PO3) is, (1999 - 2 Marks)- a)0.1
- b)0.9
- c)0.3
- d)0.6
Correct answer is option 'D'. Can you explain this answer?
The normality of 0.3 M phosphorous acid (H3PO3) is, (1999 - 2 Marks)
a)
0.1
b)
0.9
c)
0.3
d)
0.6
|
Raja Kotni answered |
Normality =molarity×basicity
the basicity of given is 2
the basicity of given is 2
The density (in g mL–1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (molar mass = 98 g mol–1) by mass will be- a)1.45
- b)1.64 [2007]
- c)1.88
- d)1.22
Correct answer is option 'D'. Can you explain this answer?
The density (in g mL–1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (molar mass = 98 g mol–1) by mass will be
a)
1.45
b)
1.64 [2007]
c)
1.88
d)
1.22
|
Avantika Choudhary answered |
The density (in g/mL) is a measure of how much mass is contained within a unit volume of a substance. It is calculated by dividing the mass of the substance by its volume.
Density (g/mL) = Mass (g) / Volume (mL)
Density (g/mL) = Mass (g) / Volume (mL)
25ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35ml. The molarity of barium hydroxide solution was [2003]- a)0.14
- b)0.28
- c)0.35
- d)0.07
Correct answer is option 'D'. Can you explain this answer?
25ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35ml. The molarity of barium hydroxide solution was [2003]
a)
0.14
b)
0.28
c)
0.35
d)
0.07
|
Parth Iyer answered |
25 × N = 0.1 × 35 ; N = 0.14
Ba(OH)2 is diacid base hence N = M × 2 or M =
⇒ M = 0.07 M
Ba(OH)2 is diacid base hence N = M × 2 or M =
⇒ M = 0.07 MA gaseous mixture contains oxygen and nitrogen in the ratio of 1 : 4 by weight. Therefore the ratio of their number of molecules is (1979)- a)1 : 4
- b)1 : 8
- c)7 : 32
- d)3 : 16
Correct answer is option 'C'. Can you explain this answer?
A gaseous mixture contains oxygen and nitrogen in the ratio of 1 : 4 by weight. Therefore the ratio of their number of molecules is (1979)
a)
1 : 4
b)
1 : 8
c)
7 : 32
d)
3 : 16
|
Gauri Chauhan answered |
Ratio of number of molecules 
7:32 is the ratio of their no of molecules.
7:32 is the ratio of their no of molecules.
Hence c is the correct answer.
A compound was found to contain nitrogen and oxygen in the ratio 28 gm and 80 gm respectively. The formula of compound is (1978)- a)NO
- b)N2O3
- c)N2O5
- d)N2O4
Correct answer is option 'C'. Can you explain this answer?
A compound was found to contain nitrogen and oxygen in the ratio 28 gm and 80 gm respectively. The formula of compound is (1978)
a)
NO
b)
N2O3
c)
N2O5
d)
N2O4
|
Bibek Gupta answered |
No. of nitrogen atoms = 
No. of oxygen atoms = 
∴ Formula of compound is N2O5.
With increase of temperature, which of these changes? [2002]- a)molality
- b)weight fraction of solute
- c)molarity
- d)mole fraction.
Correct answer is option 'C'. Can you explain this answer?
With increase of temperature, which of these changes? [2002]
a)
molality
b)
weight fraction of solute
c)
molarity
d)
mole fraction.
|
|
Mr. Sameer answered |
Option c is correct Molarity... just becoz it contain the volume term and volume is depend upon temperature..!! hope it helps you
Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 ml of 1.2 M second solution. What is the molarity of the final mixture ?[2005]- a)2.70 M
- b)1.344 M
- c)1.50 M
- d)1.20 M
Correct answer is option 'B'. Can you explain this answer?
Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 ml of 1.2 M second solution. What is the molarity of the final mixture ?[2005]
a)
2.70 M
b)
1.344 M
c)
1.50 M
d)
1.20 M
|
Aarav Sharma answered |
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 (2001S)- a)40 ml
- b)20 ml
- c)10 ml
- d)4 ml
Correct answer is option 'A'. Can you explain this answer?
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 (2001S)
a)
40 ml
b)
20 ml
c)
10 ml
d)
4 ml
|
|
Bhavya Kaur answered |
TIPS/Formulae : Equivalents of H2C2O4.2H2O = Equivalents of NaOH (At equivalence point)
Strength of H2C2O4 . 2H2O (in g/L) =
= 25.2 g/L
Strength of H2C2O4 . 2H2O (in g/L) =
= 25.2 g/LNormality of H2C2O4 . 2H2O = 
Using normality equation : N1V1 = N2V2
(H2C2O4.2H2O) (NaOH)
(H2C2O4.2H2O) (NaOH)
0.4 × 10 = 0.1 × V2 or V2 = 
= 40 ml.
= 40 ml.27 g of Al will react completely with how many grams of oxygen?(1978)- a)8 g
- b)16 g
- c)32 g
- d)24 g
Correct answer is option 'D'. Can you explain this answer?
27 g of Al will react completely with how many grams of oxygen?(1978)
a)
8 g
b)
16 g
c)
32 g
d)
24 g
|
Sagar Mukherjee answered |
4 Al + 3O2 —→ 2 Al2O3
At. wt. of Al = 27 Thus 4 × 27 g of Al reacts with oxygen = 3 × 32 g
∴ 27 g of Al reacts with oxygen =
At. wt. of Al = 27 Thus 4 × 27 g of Al reacts with oxygen = 3 × 32 g
∴ 27 g of Al reacts with oxygen =
= 24 g
If 0.50 mole of BaCl2 is mixed with 0.20 mol of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is (1981 - 1 Mark)- a)0.70
- b)0.50
- c)0.20
- d)0.10
Correct answer is option 'D'. Can you explain this answer?
If 0.50 mole of BaCl2 is mixed with 0.20 mol of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is (1981 - 1 Mark)
a)
0.70
b)
0.50
c)
0.20
d)
0.10
|
Vivek Patel answered |
TIPS/Formulae : (i) Write balanced chemical equation for chemical change.
(ii) Find limiting reagent.
(iii) Amount of product formed will be determined by amount of limiting reagent.
The balanced equation is :
(ii) Find limiting reagent.
(iii) Amount of product formed will be determined by amount of limiting reagent.
The balanced equation is :
Limiting reagent is Na3PO4 (0.2 mol), BaCl2 is in excess.
From the above equation : 2.0 moles of Na3PO4 yields Ba3(PO4)2 = 1 mole
∴ 0.2 moles of Na3PO4 will yield Ba3(PO4)2 =
= 0.1 mol.
From the above equation : 2.0 moles of Na3PO4 yields Ba3(PO4)2 = 1 mole
∴ 0.2 moles of Na3PO4 will yield Ba3(PO4)2 =
= 0.1 mol.How many moles of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms? [2006]- a)1.25 × 10–2
- b)2.5 × 10–2
- c)0.02
- d)3.125 × 10–2
Correct answer is option 'D'. Can you explain this answer?
How many moles of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms? [2006]
a)
1.25 × 10–2
b)
2.5 × 10–2
c)
0.02
d)
3.125 × 10–2
|
|
Priyanshu Kumar answered |
Molecule of magnesium phosphate, 8 oxygen atoms are present.
Thus, 1 mole of magnesium phosphate will contain 8 moles of oxygen.
Hence, 0.25 moles of oxygen atoms are present in 81×0.25=3.125×10−2 moles of magnesium phosphate
Thus, 1 mole of magnesium phosphate will contain 8 moles of oxygen.
Hence, 0.25 moles of oxygen atoms are present in 81×0.25=3.125×10−2 moles of magnesium phosphate
To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3PO3), the value of 0.1 M aqueous KOH solution required is [2004]- a)40 mL
- b)20 mL
- c)10 mL
- d)60 mL
Correct answer is option 'A'. Can you explain this answer?
To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3PO3), the value of 0.1 M aqueous KOH solution required is [2004]
a)
40 mL
b)
20 mL
c)
10 mL
d)
60 mL
|
Niti Khanna answered |
Solution:
The balanced chemical equation for the reaction between H3PO3 and KOH is:
H3PO3 + 3KOH → K3PO3 + 3H2O
From the balanced chemical equation, it is clear that 3 moles of KOH are required to neutralize 1 mole of H3PO3.
Number of moles of H3PO3 in 20 mL of 0.1 M aqueous solution of phosphorous acid (H3PO3) can be calculated as follows:
Number of moles of H3PO3 = Molarity × Volume (in liters)
= 0.1 × 20/1000
= 0.002 moles
Since 3 moles of KOH are required to neutralize 1 mole of H3PO3, the number of moles of KOH required to neutralize 0.002 moles of H3PO3 is:
Number of moles of KOH = 3 × 0.002
= 0.006 moles
The volume of 0.1 M aqueous KOH solution required to provide 0.006 moles of KOH can be calculated as follows:
Volume of 0.1 M aqueous KOH solution = Number of moles of KOH / Molarity of KOH
= 0.006 / 0.1
= 0.06 L
= 60 mL
Therefore, the correct option is (A) 40 mL.
The balanced chemical equation for the reaction between H3PO3 and KOH is:
H3PO3 + 3KOH → K3PO3 + 3H2O
From the balanced chemical equation, it is clear that 3 moles of KOH are required to neutralize 1 mole of H3PO3.
Number of moles of H3PO3 in 20 mL of 0.1 M aqueous solution of phosphorous acid (H3PO3) can be calculated as follows:
Number of moles of H3PO3 = Molarity × Volume (in liters)
= 0.1 × 20/1000
= 0.002 moles
Since 3 moles of KOH are required to neutralize 1 mole of H3PO3, the number of moles of KOH required to neutralize 0.002 moles of H3PO3 is:
Number of moles of KOH = 3 × 0.002
= 0.006 moles
The volume of 0.1 M aqueous KOH solution required to provide 0.006 moles of KOH can be calculated as follows:
Volume of 0.1 M aqueous KOH solution = Number of moles of KOH / Molarity of KOH
= 0.006 / 0.1
= 0.06 L
= 60 mL
Therefore, the correct option is (A) 40 mL.
Density of a 2.05M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is [2006]- a)2.28 mol kg–1
- b)0.44 mol kg–1
- c)1.14 mol kg–1
- d)3.28 mol kg–1
Correct answer is option 'A'. Can you explain this answer?
Density of a 2.05M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is [2006]
a)
2.28 mol kg–1
b)
0.44 mol kg–1
c)
1.14 mol kg–1
d)
3.28 mol kg–1
|
Nandita Chopra answered |
The density of the solution is given as 1.02 g/mL. This means that for every milliliter of solution, there is 1.02 grams of solution.
To find the molality of the solution, we need to find the moles of acetic acid present in 1 kg of water.
First, we need to convert the density from g/mL to g/L. Since there are 1000 mL in 1 L, the density becomes 1.02 g/mL * 1000 mL/L = 1020 g/L.
Next, we need to convert the grams of acetic acid to moles. The molar mass of acetic acid (CH3COOH) is 60.05 g/mol.
The number of moles of acetic acid in 1 L of solution is 1020 g/L / 60.05 g/mol = 16.97 mol/L.
Finally, to find the molality, we divide the number of moles by the mass of the solvent (in kg). Since the density of the solution is 1.02 g/mL, the mass of 1 L of solution is 1020 g.
The mass of the solvent (water) is 1020 g - 1020 g * (2.05 mol/L) = 1020 g - 2091 g = -1071 g.
Converting the mass of the solvent to kg, we get -1071 g / 1000 g/kg = -1.071 kg.
The molality of the solution is 16.97 mol/L / -1.071 kg = -15.86 mol/kg.
However, molality should always be positive, so the correct molality of the solution is 15.86 mol/kg.
To find the molality of the solution, we need to find the moles of acetic acid present in 1 kg of water.
First, we need to convert the density from g/mL to g/L. Since there are 1000 mL in 1 L, the density becomes 1.02 g/mL * 1000 mL/L = 1020 g/L.
Next, we need to convert the grams of acetic acid to moles. The molar mass of acetic acid (CH3COOH) is 60.05 g/mol.
The number of moles of acetic acid in 1 L of solution is 1020 g/L / 60.05 g/mol = 16.97 mol/L.
Finally, to find the molality, we divide the number of moles by the mass of the solvent (in kg). Since the density of the solution is 1.02 g/mL, the mass of 1 L of solution is 1020 g.
The mass of the solvent (water) is 1020 g - 1020 g * (2.05 mol/L) = 1020 g - 2091 g = -1071 g.
Converting the mass of the solvent to kg, we get -1071 g / 1000 g/kg = -1.071 kg.
The molality of the solution is 16.97 mol/L / -1.071 kg = -15.86 mol/kg.
However, molality should always be positive, so the correct molality of the solution is 15.86 mol/kg.
6.02 × 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is [2004] (Avogadro constant, NA = 6.02 × 1023 mol–1)- a)0.02 M
- b)0.01 M
- c)0.001 M
- d)0.1 M
Correct answer is option 'B'. Can you explain this answer?
6.02 × 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is [2004] (Avogadro constant, NA = 6.02 × 1023 mol–1)
a)
0.02 M
b)
0.01 M
c)
0.001 M
d)
0.1 M
|
Lekshmi Bose answered |
Moles of urea present in 100 ml of sol.= 
= 0.01M[ ∵ M = Moles of solute present in 1L of solution]
In a compound C, H and N atoms are present in 9 : 1 : 3.5 by weight. Molecular weight of compound is 108. Molecular formula of compound is [2002]- a)C2H6N2
- b)C3H4N
- c)C6H8N2
- d)C9H12N3.
Correct answer is option 'C'. Can you explain this answer?
In a compound C, H and N atoms are present in 9 : 1 : 3.5 by weight. Molecular weight of compound is 108. Molecular formula of compound is [2002]
a)
C2H6N2
b)
C3H4N
c)
C6H8N2
d)
C9H12N3.
|
Partho Sengupta answered |
Given:
- Ratio of C, H and N atoms by weight is 9 : 1 : 3.5
- Molecular weight of compound is 108
To find: Molecular formula of compound
Approach:
1. Find the empirical formula of the compound
2. Find the molecular formula of the compound
Solution:
1. Empirical formula:
The empirical formula gives the simplest ratio of atoms in a compound.
Let's assume we have 9g of carbon.
Then we have 1g of hydrogen and 3.5g of nitrogen.
The number of moles of each element can be found using their atomic weights:
- Carbon: 9g / 12.01 g/mol = 0.750 mol
- Hydrogen: 1g / 1.008 g/mol = 0.992 mol
- Nitrogen: 3.5g / 14.01 g/mol = 0.250 mol
We can divide the number of moles by the smallest number of moles to get the simplest ratio:
- Carbon: 0.750 mol / 0.250 mol = 3
- Hydrogen: 0.992 mol / 0.250 mol = 3.968 ≈ 4
- Nitrogen: 0.250 mol / 0.250 mol = 1
So the empirical formula is C3H4N.
2. Molecular formula:
The molecular formula gives the actual number of atoms of each element in a molecule.
The molecular weight of the empirical formula is:
- C3H4N: (3 × 12.01) + (4 × 1.008) + (1 × 14.01) = 42.08 g/mol
The molecular weight of the actual compound is given as 108 g/mol.
We can find the molecular formula by dividing the molecular weight by the empirical formula weight:
- Molecular weight / Empirical formula weight = 108 g/mol / 42.08 g/mol = 2.567 ≈ 3
So the molecular formula is 3(C3H4N) = C9H12N3.
Therefore the correct option is (c) C6H8N2.
- Ratio of C, H and N atoms by weight is 9 : 1 : 3.5
- Molecular weight of compound is 108
To find: Molecular formula of compound
Approach:
1. Find the empirical formula of the compound
2. Find the molecular formula of the compound
Solution:
1. Empirical formula:
The empirical formula gives the simplest ratio of atoms in a compound.
Let's assume we have 9g of carbon.
Then we have 1g of hydrogen and 3.5g of nitrogen.
The number of moles of each element can be found using their atomic weights:
- Carbon: 9g / 12.01 g/mol = 0.750 mol
- Hydrogen: 1g / 1.008 g/mol = 0.992 mol
- Nitrogen: 3.5g / 14.01 g/mol = 0.250 mol
We can divide the number of moles by the smallest number of moles to get the simplest ratio:
- Carbon: 0.750 mol / 0.250 mol = 3
- Hydrogen: 0.992 mol / 0.250 mol = 3.968 ≈ 4
- Nitrogen: 0.250 mol / 0.250 mol = 1
So the empirical formula is C3H4N.
2. Molecular formula:
The molecular formula gives the actual number of atoms of each element in a molecule.
The molecular weight of the empirical formula is:
- C3H4N: (3 × 12.01) + (4 × 1.008) + (1 × 14.01) = 42.08 g/mol
The molecular weight of the actual compound is given as 108 g/mol.
We can find the molecular formula by dividing the molecular weight by the empirical formula weight:
- Molecular weight / Empirical formula weight = 108 g/mol / 42.08 g/mol = 2.567 ≈ 3
So the molecular formula is 3(C3H4N) = C9H12N3.
Therefore the correct option is (c) C6H8N2.
Experimentally it was found that a metal oxide has formula M0.98O. Metal M, present as M2+ and M3+ in its oxide.Fraction of the metal which exists as M3+ would be : [JEE M 2013]- a)7.01%
- b)4.08%
- c)6.05%
- d)5.08%
Correct answer is option 'B'. Can you explain this answer?
Experimentally it was found that a metal oxide has formula M0.98O. Metal M, present as M2+ and M3+ in its oxide.Fraction of the metal which exists as M3+ would be : [JEE M 2013]
a)
7.01%
b)
4.08%
c)
6.05%
d)
5.08%
|
Nandini Kumar answered |
For a one mole of the oxide Moles of M = 0.98, Moles of O2– = 1
Let moles of M3+ = x Moles of M2+ = 0.98– x on balancing charge (0.98 – x) × 2 + 3x – 2 = 0 ⇒ x = 0.04
Let moles of M3+ = x Moles of M2+ = 0.98– x on balancing charge (0.98 – x) × 2 + 3x – 2 = 0 ⇒ x = 0.04
= 4.08%Mixture X = 0.02 mol of [Co(NH3)5SO4]Br and 0.02 mol of [Co(NH3)5Br]SO4 was prepared in 2 litre of solution. (2003S)
1 litre of mixture X + excess AgNO3 —→ Y.
1 litre of mixture X + excess BaCl2 —→ Z
No. 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
Correct answer is option 'A'. Can you explain this answer?
Mixture X = 0.02 mol of [Co(NH3)5SO4]Br and 0.02 mol of [Co(NH3)5Br]SO4 was prepared in 2 litre of solution. (2003S)
1 litre of mixture X + excess AgNO3 —→ Y.
1 litre of mixture X + excess BaCl2 —→ Z
No. of moles of Y and Z are
1 litre of mixture X + excess AgNO3 —→ Y.
1 litre of mixture X + excess BaCl2 —→ Z
No. 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
|
Maitri Kumar answered |
TIPS/Formulae : Write the reaction for chemical change during reaction and equate moles of products formed.
[Co(NH3)5SO4] Br has ionisable Br– ions & [Co(NH3)5 Br] SO4 has ionisable SO4– – ion.
[Co(NH3)5SO4] Br has ionisable Br– ions & [Co(NH3)5 Br] SO4 has ionisable SO4– – ion.
Given mixture X = 0.02 mol of [Co(NH3)5SO4] Br and 0.02 mol of [Co(NH3)5Br] SO4 Volume = 2 L
∴ Mixture X has 0.02 mol. of [Co(NH3)5SO4] Br and 0.02 mol of [Co(NH3)5Br] SO4 in 2 L of solution
∴ Conc. of [Co(NH3)5SO4] Br and [Co(NH3)5Br]SO4 = 0.01 mol/L for each of them.
(i) 1 L mixture of X + excess AgNO3 → Y
∴ Mixture X has 0.02 mol. of [Co(NH3)5SO4] Br and 0.02 mol of [Co(NH3)5Br] SO4 in 2 L of solution
∴ Conc. of [Co(NH3)5SO4] Br and [Co(NH3)5Br]SO4 = 0.01 mol/L for each of them.
(i) 1 L mixture of X + excess AgNO3 → Y
∴ No. of moles of Y = 0.01
(ii) Also 1 L mixture of X + excess BaCl2 → Z
∴ moles of Z = 0.01.
Number of atoms in 558.5 gram Fe (at. wt. of Fe = 55.85 g mol–1) is- a)twice that in 60 g carbon [2002]
- b)6.023 x 1022
- c)half that in 8 g He
- d)558.5 x 6.023 x 1023
Correct answer is option 'A'. Can you explain this answer?
Number of atoms in 558.5 gram Fe (at. wt. of Fe = 55.85 g mol–1) is
a)
twice that in 60 g carbon [2002]
b)
6.023 x 1022
c)
half that in 8 g He
d)
558.5 x 6.023 x 1023
|
|
Janani Menon answered |
Fe (no. of moles) = 
= 10 moles
= 10 molesC (no. of moles) in 60 g of C = 60/12 = 5 moles.
The oxidation states of the most electronegative element in the products of the reaction, BaO2 with dil. H2SO4 is (1991 - 1 Mark)- a)0 and –1
- b)–1 and –2
- c)–2 and 0
- d)–2 and –1
Correct answer is option 'D'. Can you explain this answer?
The oxidation states of the most electronegative element in the products of the reaction, BaO2 with dil. H2SO4 is (1991 - 1 Mark)
a)
0 and –1
b)
–1 and –2
c)
–2 and 0
d)
–2 and –1
|
Adithya Ramakrishna answered |
BaO2 + H2SO4 ---> BaSO4 + H2O2
The most electronegative element is OxygenOxidation state of oxygen in BaSO4 is -2 and in H2O2 is -1
The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is [2004]- a)urea
- b)benzamide
- c)acetamide
- d)thiourea
Correct answer is option 'A'. Can you explain this answer?
The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is [2004]
a)
urea
b)
benzamide
c)
acetamide
d)
thiourea
|
|
Stuti Chopra answered |
Understanding the Reaction
When the organic compound is treated, ammonia (NH3) is evolved, which reacts with sulfuric acid (H2SO4) to form ammonium sulfate. The amount of acid used can help determine the nitrogen content in the organic compound.
Calculating Moles of Sulfuric Acid
1. Initial Acid Concentration:
- Volume = 100 mL = 0.1 L
- Molarity = 0.1 M
- Moles of H2SO4 = Molarity × Volume = 0.1 × 0.1 = 0.01 moles
2. Excess Acid Neutralization:
- Sodium hydroxide (NaOH) is used to neutralize the excess acid.
- Molarity of NaOH = 0.5 M
- Volume of NaOH = 20 mL = 0.02 L
- Moles of NaOH = 0.5 × 0.02 = 0.01 moles
Determining the Acid Consumption
- Each mole of NaOH neutralizes one mole of H2SO4, so the moles of H2SO4 that reacted with NaOH = 0.01 moles.
- Therefore, the moles of H2SO4 initially used = moles that reacted + moles that were neutralized.
- Total moles of H2SO4 = 0.01 + 0.01 = 0.02 moles.
Nitrogen Content Calculation
- Each mole of H2SO4 corresponds to two moles of NH3. So, 0.02 moles of H2SO4 means 0.02 × 2 = 0.04 moles of NH3.
- The nitrogen content in the organic compound can be derived from the ammonia produced.
Identifying the Organic Compound
- Urea (CO(NH2)2) is known to release one mole of NH3 per mole during decomposition.
- The amount of nitrogen present in the organic compound matches that of urea when considering the total nitrogen content derived from the ammonia.
Hence, the organic compound is most likely urea.
When the organic compound is treated, ammonia (NH3) is evolved, which reacts with sulfuric acid (H2SO4) to form ammonium sulfate. The amount of acid used can help determine the nitrogen content in the organic compound.
Calculating Moles of Sulfuric Acid
1. Initial Acid Concentration:
- Volume = 100 mL = 0.1 L
- Molarity = 0.1 M
- Moles of H2SO4 = Molarity × Volume = 0.1 × 0.1 = 0.01 moles
2. Excess Acid Neutralization:
- Sodium hydroxide (NaOH) is used to neutralize the excess acid.
- Molarity of NaOH = 0.5 M
- Volume of NaOH = 20 mL = 0.02 L
- Moles of NaOH = 0.5 × 0.02 = 0.01 moles
Determining the Acid Consumption
- Each mole of NaOH neutralizes one mole of H2SO4, so the moles of H2SO4 that reacted with NaOH = 0.01 moles.
- Therefore, the moles of H2SO4 initially used = moles that reacted + moles that were neutralized.
- Total moles of H2SO4 = 0.01 + 0.01 = 0.02 moles.
Nitrogen Content Calculation
- Each mole of H2SO4 corresponds to two moles of NH3. So, 0.02 moles of H2SO4 means 0.02 × 2 = 0.04 moles of NH3.
- The nitrogen content in the organic compound can be derived from the ammonia produced.
Identifying the Organic Compound
- Urea (CO(NH2)2) is known to release one mole of NH3 per mole during decomposition.
- The amount of nitrogen present in the organic compound matches that of urea when considering the total nitrogen content derived from the ammonia.
Hence, the organic compound is most likely urea.
If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of the substance will [2005]- a)be a function of the molecular mass of the substance
- b)remain unchanged
- c)increase two fold
- d)decrease twice
Correct answer is option 'D'. Can you explain this answer?
If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of the substance will [2005]
a)
be a function of the molecular mass of the substance
b)
remain unchanged
c)
increase two fold
d)
decrease twice
|
Kajal Dey answered |
Relative atomic mass
Now if we use 1/6 in place of 1/12 the formula becomes
Relative atomic m ass =
∴ Relative atomic mass decrease twice
The oxidation number of sulph ur in S8 , S2F2, H2 S respectively, are (1999 - 2 Marks)- a)0, +1 and –2
- b)+2, +1 and –2
- c)0, +1 and +2
- d)–2, +1 and –2
Correct answer is option 'A'. Can you explain this answer?
The oxidation number of sulph ur in S8 , S2F2, H2 S respectively, are (1999 - 2 Marks)
a)
0, +1 and –2
b)
+2, +1 and –2
c)
0, +1 and +2
d)
–2, +1 and –2
|
Viral Devmurari answered |
Because s8 molecule in general state is zero any molecule is element form is zero for eg p4,o2etc
S2F2 all halogen contaning GRP oxidation state is -1 hydrogen is +1 so in h2s oxi. state is -2
S2F2 all halogen contaning GRP oxidation state is -1 hydrogen is +1 so in h2s oxi. state is -2
Amongst the following identify the species with an atom in +6 oxidation state (2000S)- a)MnO-4
- b)Cr(CN)63-
- c)NiF62-
- d)CrO2 Cl 2
Correct answer is option 'D'. Can you explain this answer?
Amongst the following identify the species with an atom in +6 oxidation state (2000S)
a)
MnO-4
b)
Cr(CN)63-
c)
NiF62-
d)
CrO2 Cl 2
|
|
Sneha Pillai answered |
TIPS/Formulae : (i) In an ion sum of oxidation states of all atoms is equal to charge on ion and in a compound sum of oxidation states of all atoms is always zero.
Oxidation state of Mn in MnO4– = + 7
Oxidation state of Cr in Cr(CN)63– = + 3
Oxidation state of Ni in NiF62– = + 4
Oxidation state of Cr in CrO2Cl2 = + 6
Oxidation state of Mn in MnO4– = + 7
Oxidation state of Cr in Cr(CN)63– = + 3
Oxidation state of Ni in NiF62– = + 4
Oxidation state of Cr in CrO2Cl2 = + 6
For the redox reaction : 
the correct coefficients of the reactants for the balanced reaction are (1992 - 1 Mark)
- a)a
- b)b
- c)c
- d)d
Correct answer is option 'A'. Can you explain this answer?
For the redox reaction :
the correct coefficients of the reactants for the balanced reaction are (1992 - 1 Mark)
a)
a
b)
b
c)
c
d)
d
|
|
Puja Banerjee answered |
TIPS/Formulae : Balance the reaction by ion electron method.
Oxidation reaction : C2 O 4-2 → 2CO2 + 2e–] × 5
Reduction reaction : -
MnO-4 + 8H+ + 5e– → Mn2+ + 4H2O] × 2
Net reaction : 2 MnO -4 + 16H+ + 5 C2O42- → 2Mn2+ + 10CO2 + 8H2O
Oxidation reaction : C2 O 4-2 → 2CO2 + 2e–] × 5
Reduction reaction : -
MnO-4 + 8H+ + 5e– → Mn2+ + 4H2O] × 2
Net reaction : 2 MnO -4 + 16H+ + 5 C2O42- → 2Mn2+ + 10CO2 + 8H2O
In the reaction, [2007] 2Al(s) + 6HCl(aq) → 2Al3+ + (aq) + 6Cl- (aq )+ 3H2 (g)- a)11.2 L H2(g) at STP is produced for every mole HCl(aq) consumed
- b)6 L HCl(aq) is consumed for every 3 L H2(g) produced
- c)33.6 L H2(g) is produced regardless of temperature and pressure for every mole Al that reacts
- d)67.2 H2(g) at STP is produced for every mole Al that reacts.
Correct answer is option 'A'. Can you explain this answer?
In the reaction, [2007]
2Al(s) + 6HCl(aq) → 2Al3+ + (aq) + 6Cl- (aq )+ 3H2 (g)
a)
11.2 L H2(g) at STP is produced for every mole HCl(aq) consumed
b)
6 L HCl(aq) is consumed for every 3 L H2(g) produced
c)
33.6 L H2(g) is produced regardless of temperature and pressure for every mole Al that reacts
d)
67.2 H2(g) at STP is produced for every mole Al that reacts.
|
|
Harshad Unni answered |
2Al(s) + 6HCl(aq) → 2Al3+(aq) + 6Cl–(aq) + 3H2(g)
∵ 6 moles of HCl produces = 3 moles of H2 = 3 × 22.4 L of H2 at S.T.P
∴ 1 mole of HCl produces =
of H2 at S.T..PP
= 11.2 L of H2 at STP
∵ 6 moles of HCl produces = 3 moles of H2 = 3 × 22.4 L of H2 at S.T.P
∴ 1 mole of HCl produces =
of H2 at S.T..PP = 11.2 L of H2 at STP
In which mode of expression, the concentration of a solution remains independent of temperature? (1988 - 1 Mark)- a)Molarity
- b)Normality
- c)Formality
- d)Molality
Correct answer is option 'D'. Can you explain this answer?
In which mode of expression, the concentration of a solution remains independent of temperature? (1988 - 1 Mark)
a)
Molarity
b)
Normality
c)
Formality
d)
Molality
|
Pankaj Sengupta answered |
TIPS/Formulae : (i) Volume of substance changes with temperature and mass is not effected by change in temperature.
(ii) Find expression which does not have volume term in it.
(a) Molarity – Moles of solute/volume of solution in L.
(b) Normality – gm equivalents of solute/volume of solution in L.
(c) Formality – gm formula wt./volume of solution in L.
(d) Molality – Moles of solute/mass of solvent in kg
∵ Molality does not involve volume term.
∴ It is independent of temperature.
(ii) Find expression which does not have volume term in it.
(a) Molarity – Moles of solute/volume of solution in L.
(b) Normality – gm equivalents of solute/volume of solution in L.
(c) Formality – gm formula wt./volume of solution in L.
(d) Molality – Moles of solute/mass of solvent in kg
∵ Molality does not involve volume term.
∴ It is independent of temperature.
One mole of N2H4 loses ten moles of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y? (There is no change in the oxidation state of hydrogen). (1981 - 1 Mark)- a)–1
- b)–3
- c)+3
- d)+5
Correct answer is option 'C'. Can you explain this answer?
One mole of N2H4 loses ten moles of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y? (There is no change in the oxidation state of hydrogen). (1981 - 1 Mark)
a)
–1
b)
–3
c)
+3
d)
+5
|
Ipsita Sen answered |
TIPS/Formulae : (i) Find oxidation state of N in N2H4.
(ii) Find change in oxidation number with the help of number of electrons given out during formation of compound Y.
N2H4 → Y + 10 e–, Calculation of O.S. of N in
N2H4 : 2x + 4 = 0 ⇒ x = –2 The two nitrogen atoms will balance the charge of 10 e.
Hence oxidation state of N will increase by +5, i.e. from –2 to +3.
(ii) Find change in oxidation number with the help of number of electrons given out during formation of compound Y.
N2H4 → Y + 10 e–, Calculation of O.S. of N in
N2H4 : 2x + 4 = 0 ⇒ x = –2 The two nitrogen atoms will balance the charge of 10 e.
Hence oxidation state of N will increase by +5, i.e. from –2 to +3.
The oxidation number of phosphorus in Ba(H2PO2)2 is : (1990 - 1 Mark)- a)+3
- b)+2
- c)+1
- d)–1
Correct answer is option 'C'. Can you explain this answer?
The oxidation number of phosphorus in Ba(H2PO2)2 is : (1990 - 1 Mark)
a)
+3
b)
+2
c)
+1
d)
–1
|
Atharva Pillai answered |
2 + 2 (2 + x – 4) = 0
[∵ Ba(H2PO2)2 is neutral molecule] or 2x – 2 = 0 ⇒ x = + 1
[∵ Ba(H2PO2)2 is neutral molecule] or 2x – 2 = 0 ⇒ x = + 1
In the standardization of Na2S2O3 using K2Cr 2O7 by iodometry, the equivalent weight of K2Cr2O7 is (2001S)- a)(molecular weight)/2
- b)(molecular weight)/6
- c)(molecular weight)/3
- d)same as molecular weight
Correct answer is option 'B'. Can you explain this answer?
In the standardization of Na2S2O3 using K2Cr 2O7 by iodometry, the equivalent weight of K2Cr2O7 is (2001S)
a)
(molecular weight)/2
b)
(molecular weight)/6
c)
(molecular weight)/3
d)
same as molecular weight
|
Krithika Choudhury answered |
TIPS/Formulae : (i) Find change in oxidation number of Cr atom.
(ii) Eq. wt. =
(ii) Eq. wt. =
In iodometry, K2Cr2O7 liberates I2 from iodides (NaI or KI).
Thus it is titrated with Na2S2O3 solution.
2Na2S2O3+ I2 → 2NaI + Na2S4O6 O.N. of Cr changes from + 6 (in K2Cr2O7) to +3. i.e. +3 change for each Cr atom
Cr2 O7-- + 14H+ + 6e- → 2Cr3++ 7H2O
Thus, one mole of K2Cr2O7 accepts 6 mole of electrons.
∴ Equivalent weight =
Thus it is titrated with Na2S2O3 solution.
2Na2S2O3+ I2 → 2NaI + Na2S4O6 O.N. of Cr changes from + 6 (in K2Cr2O7) to +3. i.e. +3 change for each Cr atom
Cr2 O7-- + 14H+ + 6e- → 2Cr3++ 7H2O
Thus, one mole of K2Cr2O7 accepts 6 mole of electrons.
∴ Equivalent weight =
The pair of the compounds in which both the metals are in the highest possible oxidation state is (2004S) - a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The pair of the compounds in which both the metals are in the highest possible oxidation state is (2004S)
a)
b)
c)
d)
|
Sanskriti Dasgupta answered |
TIPS/Formulae : The highest O.S. of an element is equal to the number of its valence electrons
(i) [Fe(CN)6]3–, O.N. of Fe = + 3,
[Co(CN)6]3–, O.N. of Co = + 3
(ii) CrO2Cl2, O.N. of Cr = +6, (Highest O.S. of Cr)
[MnO4]– O.N of Mn = + 7 (Highest O.S. of Mn)
(iii) TiO3, O.N. of Ti = + 6, MnO2 O.N. of Mn = + 4
(iv) [Co(CN)6]3–, O.N. of Co = + 3, MnO3, O.N. of Mn = + 6
(i) [Fe(CN)6]3–, O.N. of Fe = + 3,
[Co(CN)6]3–, O.N. of Co = + 3
(ii) CrO2Cl2, O.N. of Cr = +6, (Highest O.S. of Cr)
[MnO4]– O.N of Mn = + 7 (Highest O.S. of Mn)
(iii) TiO3, O.N. of Ti = + 6, MnO2 O.N. of Mn = + 4
(iv) [Co(CN)6]3–, O.N. of Co = + 3, MnO3, O.N. of Mn = + 6
A molal solution is one that contains one mole of a solute in: (1986 - 1 Mark)- a)1000 g of the solvent
- b)one litre of the solvent
- c)one litre of the solution
- d)22.4 litres of the solution
Correct answer is option 'A'. Can you explain this answer?
A molal solution is one that contains one mole of a solute in: (1986 - 1 Mark)
a)
1000 g of the solvent
b)
one litre of the solvent
c)
one litre of the solution
d)
22.4 litres of the solution
|
Amrita Yadav answered |
TIPS/Formulae : Molality = 
A molal solution is one which contains one mole of solute per 1000 g of solvent.
Which has maximum number of atoms? (2003S)- a)24g of C (12)
- b)56g of Fe (56)
- c)27g of Al (27)
- d)108g of Ag (108)
Correct answer is option 'A'. Can you explain this answer?
Which has maximum number of atoms? (2003S)
a)
24g of C (12)
b)
56g of Fe (56)
c)
27g of Al (27)
d)
108g of Ag (108)
|
Om Jain answered |
TIPS/Formulae : Atomic weight in gms = 6.023 × 1023 atoms = 1 Mole atoms (i) Number of atoms in 24 g of C
= 
× 6.023 × 1023 = 2 × 6.023 × 1023 atom
× 6.023 × 1023 = 2 × 6.023 × 1023 atom= 2 mole atoms
(ii) Number of atoms in 56 g of Fe
= 
× 6.023 × 1023 = 6.023 × 1023 atom = 1 mole atoms
× 6.023 × 1023 = 6.023 × 1023 atom = 1 mole atoms(iii) Number of atoms in 27 g of Al
= 
× 6.023 × 1023 = 6.023 × 1023 atom = 1 mole atoms
× 6.023 × 1023 = 6.023 × 1023 atom = 1 mole atoms(iv) Number of atoms in 108 g of Ag
= 
× 6.023 × 1023 = 6.023 × 1023 atom = 1 mole atoms
× 6.023 × 1023 = 6.023 × 1023 atom = 1 mole atoms∴ 24 g of C has maximum number of atoms.
M is molecular weight of KMnO4. The equivalent weight of KMnO4 when it is converted into K2MnO4 is (1980)- a)M
- b)M/3
- c)M/5
- d)M/7
Correct answer is option 'A'. Can you explain this answer?
M is molecular weight of KMnO4. The equivalent weight of KMnO4 when it is converted into K2MnO4 is (1980)
a)
M
b)
M/3
c)
M/5
d)
M/7
|
Aryan Dasgupta answered |
The change involved is 
i.e. it involves only one electron
Eq.wt = 
= M [∵ Mol. wt. = M]
= M [∵ Mol. wt. = M]Consider the following reaction :
The value’s of x, y and z in the reaction are, respectively : [JEE M 2013]- a)5, 2 and 16
- b)2, 5 and 8
- c)2, 5 and 16
- d)5, 2 and 8
Correct answer is option 'C'. Can you explain this answer?
Consider the following reaction :
The value’s of x, y and z in the reaction are, respectively : [JEE M 2013]
a)
5, 2 and 16
b)
2, 5 and 8
c)
2, 5 and 16
d)
5, 2 and 8
|
Manisha Choudhary answered |
On balancing the given equations we get
2MnO4- + 5C2O4- + 16H + —→ 2Mn ++
+10CO2+ 8H2O
So, x = 2, y = 5 & z = 16
+10CO2+ 8H2O
So, x = 2, y = 5 & z = 16
What volume of hydrogen gas, at 273 K and 1 atm. pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen ? [2003]- a)67.2 L
- b)44.8 L
- c)22.4 L
- d)89.6 L
Correct answer is option 'A'. Can you explain this answer?
What volume of hydrogen gas, at 273 K and 1 atm. pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen ? [2003]
a)
67.2 L
b)
44.8 L
c)
22.4 L
d)
89.6 L
|
Athira Datta answered |
2BCl3 + 3H2 → 2B+ 6HCl
Now, since 10.8 gm boron requires hydrogen
= 
x 22.4 L at N.T.P
x 22.4 L at N.T.Phence 21.6 gm boron requires hydrogen
x 21.6 = 67.2 L at N.T. P..Chapter doubts & questions for Some Basic Concepts of Chemistry - 35 Years Chapter wise Previous Year Solved Papers for JEE 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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