Question 1

A 100 mL sample of a urea solution contains 6.02 × 10²⁰ molecules of urea. What is the molarity of the solution? (Take Avogadro constant, N_A = 6.02 × 10²³ mol^−1.)

Accepted Answer

0.01 M. 6.02e20 / 6.02e23 = 1e-3 mol of urea in 0.100 L, so molarity = 0.001/0.1 = 0.01 M. Stored answer 0.001 M confuses moles with molarity.

Question 2

When 4.9 g of H2SO4 reacts with NaCl, the products formed are 6 g of sodium hydrogen sulphate and 1.825 g of HCl. What mass of NaCl was decomposed?

Accepted Answer

2.925 g. H2SO4 (4.9 g) + NaCl = NaHSO4 (6 g) + HCl (1.825 g); by conservation NaCl = 7.825 - 4.9 = 2.925 g (= 0.05 mol). Stored 4.65 g is wrong.

Question 3

Which of the following has the least mass?

Accepted Answer

0.1 g of silver. 0.2 mol H2 = 0.4 g; 6.023e22 N2 = 0.1 mol = 2.8 g; 0.1 g Ag = 0.1 g; 0.1 mol O2 = 3.2 g. Least is 0.1 g of silver, not the hydrogen.

Question 4

What amount of magnesium phosphate, Mg3(PO4)2, in moles is needed to provide 0.25 mole of oxygen atoms?

Accepted Answer

3.125 × 10^−2. Each Mg3(PO4)2 contains 8 O atoms, so moles = 0.25/8 = 0.03125 = 3.125e-2. Stored 2.5e-2 is wrong.

Question 5

Haemoglobin has 0.33% iron by mass. If its molecular mass is about 67200 and the atomic mass of iron is 56, how many iron atoms are present in a single haemoglobin molecule?

Accepted Answer

4. The percentage of iron by mass in haemoglobin indicates that for every 100 grams of haemoglobin, there are 0.33 grams of iron. Given the molecular mass of haemoglobin is approximately 67200 g/mol, the mass of iron in one mole of haemoglobin is calculated to be 0.33% of 67200, which equals about 221 grams. Dividing this by the atomic mass of iron (56 g/mol) shows that there are approximately 4 iron atoms in each haemoglobin molecule.

Question 6

For the reaction 2Al(s) + 6HCl(aq) → 2Al³+(aq) + 6Cl^−(aq) + 3H₂(g), which statement is correct?

Accepted Answer

11.2 L of H₂(g) at STP is formed for each mole of HCl(aq) used up. 2Al->3H2 means 1 mol Al gives 1.5 mol H2 = 33.6 L (not 67.2). The correct true statement is option 0: 6 HCl -> 3 H2, so 1 mol HCl gives 0.5 mol H2 = 11.2 L at STP.

Question 7

Commercial concentrated sulphuric acid contains 95% H₂SO₄ by mass. If its density is 1.834 g cm⁻³, what is the molarity of the solution?

Accepted Answer

17.8 M. The correct option is right because the molarity is determined by the concentration of the solute in the solution, which is calculated using the density and mass percentage of the acid, leading to a result of 17.8 M.

Question 8

When 50 mL of a 16.9% AgNO₃ solution is combined with 50 mL of a 5.8% NaCl solution, what mass of precipitate is produced? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)

Accepted Answer

7 g. The correct option is 7 g because the reaction between AgNO₃ and NaCl produces AgCl as a precipitate, and stoichiometric calculations based on the concentrations and volumes of the solutions show that 7 g of AgCl is formed.

Question 9

On complete combustion, a gaseous hydrocarbon produces 0.72 g of water and 3.08 g of carbon dioxide. What is the empirical formula of this hydrocarbon?

Accepted Answer

C7H8. CO2 3.08/44 = 0.07 mol C; H2O 0.72/18 = 0.04 mol -> 0.08 mol H. C:H = 0.07:0.08 = 7:8 -> C7H8. Stored C3H4 is wrong.

Question 10

A sample of washing soda has the following composition by mass:

- Na₂CO₃: molecular mass 106.0, mass percentage 84.8
- NaHCO₃: molecular mass 84.0, mass percentage 8.4
- NaCl: molecular mass 58.5, mass percentage 6.8

If 1 kg of this sample is treated completely with excess HCl, the amoun

Accepted Answer

9 mol of CO₂. Na2CO3 848/106 = 8 mol -> 8 CO2; NaHCO3 84/84 = 1 mol -> 1 CO2; NaCl gives none. Total 9 mol CO2. Stored 17 is wrong.

Question 11

A nitrogen–hydrogen gaseous compound has hydrogen making up 12.5% of its mass. Its density with respect to hydrogen is 16. What is the molecular formula of this compound?

Accepted Answer

N₂H₄. The correct option, N₂H₄, is derived from the given mass percentage of hydrogen and the density ratio, which indicates that the compound contains two nitrogen atoms for every four hydrogen atoms, aligning with the calculated molecular formula.

Question 12

When 100 mL of a 20.8% BaCl₂ solution is mixed with 50 mL of a 9.8% H₂SO₄ solution, the mass of BaSO₄ produced will be: (Ba = 137, Cl = 35.5, S = 32, H = 1, O = 16)

Accepted Answer

11.65 g. BaCl2 = 20.8 g/208 = 0.1 mol; H2SO4 = 4.9 g/98 = 0.05 mol (limiting). BaSO4 = 0.05 x 233 = 11.65 g. Stored 30.6 g is wrong.

Question 13

A 2 g sample containing only CO and CO₂ is treated with excess I₂O₅. If 2.54 g of I₂ is obtained, what is the mass percentage of CO₂ in the original sample?

Accepted Answer

30. I2 = 2.54/254 = 0.01 mol; 5CO -> 1 I2, so CO = 0.05 mol = 1.4 g, CO2 = 0.6 g -> 30% CO2. Stored 70 is wrong.

Question 14

A sample of a gas has a mass of 7.5 g and occupies 5.6 L at STP. Which gas is it?

Accepted Answer

NO. 5.6 L at STP = 0.25 mol; molar mass = 7.5/0.25 = 30 g/mol = NO. Stored CO (28 g/mol) is wrong.

Question 15

How much oxygen is needed to completely burn 2.8 kg of ethylene?

Accepted Answer

9.6 kg. To completely burn ethylene (C2H4), a specific stoichiometric amount of oxygen is required. The balanced chemical equation shows that for every mole of ethylene, three moles of oxygen are needed, which translates to approximately 9.6 kg of oxygen for 2.8 kg of ethylene.

Question 16

A 3 L mixture containing propane (C₃H₈) and butane (C₄H₁₀) is completely burned at 25°C. If the combustion yields 10 L of carbon dioxide, determine the volume ratio of propane to butane in the original mixture.

Accepted Answer

2:1. Let propane=x, butane=y L. x+y=3 and 3x+4y=10 give y=1, x=2, so propane:butane = 2:1. Stored 1:2 is inverted.

Question 17

An organic substance is found to have 49.3% carbon and 6.84% hydrogen by mass. If its vapour density is 73, what is its molecular formula?

Accepted Answer

C₆H₁₀O₄. C:H:O = 49.3/12 : 6.84/1 : 43.86/16 = 1.5:2.5:1 = 3:5:2 -> empirical C3H5O2 (73); MW 146 doubles it to C6H10O4. Stored C4H10O2 (90) is wrong.

Question 18

Which of the following groups does not consist of isoelectronic species?

Accepted Answer

SO₃²⁻, CO₃²⁻, NO₃⁻. SO3^2- has 42 electrons while CO3^2- and NO3^- have 32 each, so set (option 1) is NOT isoelectronic. The stored set PO4^3-/SO4^2-/ClO4^- all have 50 electrons (isoelectronic), so it is not the answer.

Question 19

The bond dissociation energy for one mole of Cl–Cl bonds in Cl₂ is 242 kJ mol⁻¹. What is the maximum wavelength of radiation that can just rupture one Cl–Cl bond? (Take c = 3 × 10⁸ m s⁻¹ and Nₐ = 6.02 × 10²³ mol⁻¹.)

Accepted Answer

494 nm. Per-bond energy = 242000/6.02e23 = 4.02e-19 J; lambda = (6.626e-34 x 3e8)/4.02e-19 = 4.94e-7 m = 494 nm. The stored 594 nm is wrong.

Question 20

When carbon dioxide gas is passed over red-hot coke, part of it is converted into carbon monoxide. If 0.5 L of CO₂ is passed over red-hot coke and the total volume of the resulting gaseous mixture becomes 700 mL, what is the composition of the gas mixture at STP?

Accepted Answer

CO₂ = 300 mL; CO = 400 mL. If x mL CO2 reacts, total gas = (500 - x) + 2x = 500 + x = 700, so x = 200. Remaining CO2 = 300 mL and CO = 400 mL, not 200/500.

Question 21

A 5 L mixture containing methane and propane is completely burned at 0°C and 1 atm. If 16 L of oxygen, measured at the same temperature and pressure, is used up, what is the heat evolved in this combustion, in kJ? (Given: ΔHcomb(CH₄) = 890 kJ mol⁻¹ and ΔHcomb(C₃H₈) = 2220 kJ mol⁻¹)

Accepted Answer

317. The correct option is derived from calculating the moles of methane and propane in the mixture based on the stoichiometry of their combustion reactions with oxygen. By determining the amount of each gas burned and applying their respective enthalpy changes of combustion, the total heat evolved can be accurately calculated, leading to the correct answer of 317 kJ.

Question 22

For the combustion reaction of ethane, C2H6(g) + nO2(g) → CO2(g) + H2O(l), what is the ratio of the stoichiometric coefficients of carbon dioxide to water in the balanced equation?

Accepted Answer

2: 3. Balanced ethane combustion gives 2 CO2 and 3 H2O, so the CO2:H2O ratio is 2:3, not 1:1.

Question 23

How many moles of H2 are present in 0.224 L of hydrogen gas measured at STP (273 K and 1 atm)?

Accepted Answer

0.01. At standard temperature and pressure (STP), one mole of any gas occupies 22.4 liters. Therefore, to find the number of moles in 0.224 liters, you divide 0.224 L by 22.4 L/mol, resulting in 0.01 moles of H2.

Question 24

In Dumas’ method for nitrogen estimation, 0.25 g of an organic substance yields 40 mL of nitrogen gas collected at 300 K and 725 mm pressure. If the vapour pressure of water at 300 K is 25 mm, the nitrogen percentage in the compound is:

Accepted Answer

16.76. The correct option is derived by first adjusting the pressure of the nitrogen gas to account for the vapor pressure of water, then using the ideal gas law to calculate the number of moles of nitrogen produced. Finally, the percentage of nitrogen in the original compound is calculated based on the mass of the organic substance and the moles of nitrogen obtained.

Question 25

An organic compound sample weighing 0.32 g gives 0.233 g of BaSO₄ on analysis. Taking atomic masses Ba = 137 and S = 32, what is the percentage of sulphur in the compound?

Accepted Answer

10.0. S = 0.233 x 32/233 = 0.032 g; %S = 0.032/0.32 x 100 = 10.0%. The stored 1.0% is off by a factor of ten.

Question 26

An organic compound has the composition 38.8% carbon, 16.0% hydrogen, and 45.2% nitrogen by mass. Which molecular formula matches this compound?

Accepted Answer

CH₃NH₂. Moles C:H:N = 38.8/12 : 16/1 : 45.2/14 = 1 : 5 : 1, giving CH5N = CH3NH2 (which matches the percentages exactly). The stored CH2(NH2)2 does not fit.

Question 27

What mass of concentrated nitric acid solution is required to make 250 mL of 2.0 M HNO3, if the concentrated acid contains 70% HNO3 by mass?

Accepted Answer

45.0 g concentrated HNO3 solution. Need 0.5 mol HNO3 = 31.5 g; at 70% w/w the solution mass = 31.5/0.70 = 45 g, not 70 g.

Question 28

A sample of 25.3 g of sodium carbonate, Na2CO3, is dissolved in water and the total volume is made up to 250 mL. Assuming complete dissociation, what are the molar concentrations of Na+ and CO3²− ions, respectively? (Molar mass of Na2CO3 = 106 g mol−1)

Accepted Answer

1.910 M and 0.955 M. The correct option is right because when sodium carbonate (Na2CO3) dissolves, it dissociates into two sodium ions (Na+) and one carbonate ion (CO3²−). Therefore, the concentration of Na+ ions is twice that of CO3²− ions, leading to 1.910 M for Na+ and 0.955 M for CO3²− in a 250 mL solution.

Question 29

For the reaction N2(g) + 3H2(g) → 2NH3(g), what is the correct relation between the rate of change of ammonia concentration and the rate of change of hydrogen concentration?

Accepted Answer

d[NH3]/dt = -2 d[H2]/dt / 3. For N2 + 3H2 -> 2NH3: (1/2) d[NH3]/dt = -(1/3) d[H2]/dt. Therefore d[NH3]/dt = -(2/3) d[H2]/dt, i.e. d[NH3]/dt = -2 d[H2]/dt / 3.

Question 30

Polyethylene is prepared from calcium carbide through the sequence shown below:
CaC2 + 2H2O → Ca(OH)2 + C2H2
C2H2 + H2 → C2H4
nC2H4 → (–CH2–CH2–)n
What mass of polyethylene can be produced from 64.1 kg of CaC2?

Accepted Answer

28 kg. The reaction shows that 1 mole of calcium carbide (CaC2) produces 1 mole of acetylene (C2H2), which then converts to polyethylene (–CH2–CH2–)n. Given the molar mass of CaC2 and the stoichiometry of the reactions, 64.1 kg of CaC2 yields enough C2H4 to produce 28 kg of polyethylene.

Question 31

For the reaction 2Al(s) + 6HCl(aq) → 2Al³+(aq) + 6Cl^-(aq) + 3H2(g), which statement is correct?

Accepted Answer

11.2 L of H2(g) at STP is formed for each mole of HCl(aq) used up.. From the stoichiometry 6 HCl -> 3 H2, each mole of HCl yields 0.5 mol H2 = 0.5 x 22.4 = 11.2 L at STP. (1 mol Al gives 1.5 mol = 33.6 L H2, not 67.2 L, so that option is wrong.)

Question 32

What is the molality of a urea solution prepared by dissolving 0.0100 g of urea, (NH2)2CO, in 0.3000 dm³ of water at STP?

Accepted Answer

5.55 × 10⁻⁴ m. Moles urea = 0.0100/60 = 1.667e-4 mol. Mass of water = 0.300 dm^3 ~ 0.300 kg. Molality = 1.667e-4 / 0.300 = 5.55e-4 m.

Question 33

On complete combustion, a gaseous hydrocarbon produces 0.72 g of water and 3.08 g of carbon dioxide. What is its empirical formula?

Accepted Answer

C7H8. CO2: 3.08/44 = 0.07 mol C. H2O: 0.72/18 = 0.04 mol -> 0.08 mol H. C:H = 0.07:0.08 = 7:8, so empirical formula is C7H8.

Question 34

In a healthy adult human body, the most abundant elements by mass are oxygen (61.4%), carbon (22.9%), hydrogen (10.0%), and nitrogen (2.6%). If every 1H atom in a 75 kg person were replaced by 2H atoms, by how much would the person's mass increase?

Accepted Answer

7.5 kg. Replacing each hydrogen atom with two deuterium atoms effectively doubles the mass contribution from hydrogen. Since hydrogen makes up about 10% of a 75 kg person, this results in an increase of approximately 7.5 kg.

Question 35

A 1 g sample of a carbonate having the formula M2CO3 is treated with excess hydrochloric acid. If 0.01186 mol of carbon dioxide is evolved, what is the molar mass of M2CO3 in g mol⁻¹?

Accepted Answer

84.3. Each M2CO3 gives one CO2, so moles M2CO3 = 0.01186. Molar mass = 1 g / 0.01186 mol = 84.3 g/mol.

Question 36

In an organic compound of formula CxHyOz, the mass percentages of carbon and hydrogen are in the ratio 6:1. Also, a single molecule of this compound contains only half the oxygen needed for complete combustion of one molecule of CxHy to carbon dioxide and water. What is the empirical formu

Accepted Answer

C2H4O3. 12x/y = 6 gives y = 2x, so the hydrocarbon part is CxH2x. Complete combustion needs 3x O atoms (2x for CO2 + x for H2O); half of that is 1.5x, so z = 1.5x. With x = 2 the empirical formula is C2H4O3.

Question 37

For a reaction, N2(g) + 3H2(g) → 2NH3(g); identify dihydrogen (H2) as a limiting reagent in the following reaction mixtures.

Accepted Answer

56 g of N2 + 10 g of H2. In the reaction, 1 mole of N2 reacts with 3 moles of H2. In option A, the amount of H2 available (10 g or about 5 moles) is insufficient to fully react with the available N2 (56 g or about 2 moles), making H2 the limiting reagent.

Question 38

Amongst the following statements, that which was not proposed by Dalton was:

Accepted Answer

When gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T & P.. This statement, while related to gas behavior, was actually proposed by Avogadro, not Dalton. Dalton's atomic theory focused on the indivisibility of atoms and their mass properties, but did not specifically address the volume ratios of gases.

Question 39

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1: 4. The ratio of number of their molecules is:

Accepted Answer

7: 32. For mass ratio O:N = 1:4, moles of O2 = 1/32 and moles of N2 = 4/28 = 1/7. The molecule ratio = (1/32):(1/7) = 7:32.

Question 40

For which of the following properties would the structural isomers C2H5OH and CH3OCH3 be expected to show identical values, assuming ideal behavior?

Accepted Answer

Gaseous densities at the same temperature and pressure. C2H5OH and CH3OCH3 have the same molecular formula and therefore the same molar mass. At the same temperature and pressure, gaseous density = PM/RT depends only on molar mass, so their gaseous densities are identical.

Question 41

The molarity of a solution obtained by mixing 750 mL of 0.5 M HCl with 250 mL of 2 M HCl will be:

Accepted Answer

0.875 M. To find the molarity of the mixed solution, we calculate the total moles of HCl from each solution and divide by the total volume of the mixture. The moles from the first solution (0.5 M) and the second solution (2 M) are combined, and the total volume is 1000 mL, resulting in a final molarity of 0.875 M.

Question 42

Synthesis of each molecule of glucose in photosynthesis involves:

Accepted Answer

18 molecules of ATP. Fixing 6 CO2 to make one glucose in the Calvin cycle consumes 18 molecules of ATP (and 12 NADPH). So the answer is 18 ATP.

Question 43

29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compou

Accepted Answer

23.7. The correct option is right because the amount of nitrogen is calculated based on the neutralization of the excess acid after ammonia absorption. By determining the moles of HCl initially present and subtracting the moles neutralized by NaOH, the moles of ammonia (and thus nitrogen) can be found, leading to the percentage of nitrogen in the compound.

Question 44

For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20 mL of M/10 sodium hydroxide for complete neutralization. The percentage of nitrogen in the compo

Accepted Answer

10%. The correct option is 10% because the amount of ammonia produced from the organic compound can be calculated from the neutralization of the unreacted sulfuric acid, which indicates the nitrogen content. By determining the moles of ammonia and relating it to the mass of the organic compound, the percentage of nitrogen can be accurately derived.

Question 45

A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2. The empirical formula of the hydrocarbon is -

Accepted Answer

C7H8. The correct option, C7H8, is determined by calculating the moles of carbon and hydrogen from the combustion products. The mass of CO2 indicates the amount of carbon, while the mass of water indicates the amount of hydrogen, leading to a ratio that corresponds to the empirical formula C7H8.

Question 46

A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is -

Accepted Answer

5. The increase in molecular mass from 180 to 390 indicates that 210 mass units are added during acylation. Since the acylation of an amino group with CH3COCl adds 42 mass units per amino group, dividing 210 by 42 gives 5, indicating there are 5 amino groups in the original compound.

Question 47

Experimentally it was found that a metal oxide has formula M0.98O. Metal M is present as M2+ and M3+ in its oxide. Fraction of metal which exists as M3+ would be -

Accepted Answer

4.08%. To find the fraction of metal M that exists as M3+, we can set up a balance of charges in the oxide formula M0.98O. Given that M is present as both M2+ and M3+, we can express the total amount of metal in terms of its oxidation states and solve for the fraction of M3+, leading to the conclusion that 4.08% of the metal is in the M3+ state.

Question 48

Synthesis of each molecule of glucose in photosynthesis involves -

Accepted Answer

18 molecules of ATP. The synthesis of one molecule of glucose during photosynthesis requires a total of 18 ATP molecules, as it involves multiple cycles of the Calvin cycle where ATP is used to convert carbon dioxide into glucose.

Question 49

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1: 4. The ratio of number of their molecule is -

Accepted Answer

7: 32. To find the ratio of the number of molecules, we first need to convert the mass ratio into a mole ratio using the molar masses of oxygen (16 g/mol) and nitrogen (28 g/mol). Given the mass ratio of oxygen to nitrogen is 1:4, we can calculate the number of moles of each gas and find that the ratio of their molecules is 7:32.

Question 50

For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20 mL of M/10 sodium hydroxide for complete neutralization. The percentage of nitrogen in the compo

Accepted Answer

10%. The correct option is 10% because the amount of ammonia produced from the digestion of the organic compound can be calculated based on the neutralization of the unreacted sulfuric acid with sodium hydroxide. This calculation reveals that the nitrogen content in the original sample corresponds to 10% of its weight.

JEE Main Chemistry: Some Basic Concepts of Chemistry questions with solutions

Questions