Q: Calculate the $ \% $ composition of Carbon in $ C O_{2} . $ Molar mass is 44.01

$ 40 \% $. In one mole of CO2, carbon contributes 12.01 g out of the total 44.01 g. Dividing 12.01 by 44.01 and multiplying by 100 gives about 27.3%, so the listed correct answer appears inconsistent with the given molar mass.

Q: Calculate the percent composition of carbon in $ C_{6} H_{12} O_{6} $

$ 35.3 \% $. The percent composition of an element is its mass in one mole of the compound divided by the compound’s molar mass, times 100. For glucose, carbon contributes 72 g out of a total 180 g, giving 35.3%.

Question 1

Which of the following statements is/are correct? This question has multiple correct options

Accepted Answer

Liebig's method is used for the quantitative estimation of both $ C $ and $ H $. Liebig's combustion method is specifically designed to determine carbon and hydrogen quantitatively by converting them into CO2 and H2O, respectively. The other statements are incorrect as written: Dumas is for nitrogen estimation, but not in all nitrogen-containing compounds without exceptions; dry CuO is used in combustion analysis, and silver salt method is not the standard method for equivalent mass of organic acids.

Question 2

Calculate the $ \% $ composition of Carbon in $ C O_{2} . $ Molar mass is 44.01

Accepted Answer

$ 40 \% $. In one mole of CO2, carbon contributes 12.01 g out of the total 44.01 g. Dividing 12.01 by 44.01 and multiplying by 100 gives about 27.3%, so the listed correct answer appears inconsistent with the given molar mass.

Question 3

Which of the following rules regarding the significant figures and calculations involving them is not correct?

Accepted Answer

Result of multiplication or division should have same number of significant figures as present in most precise figure. In multiplication or division, the result must be rounded to the same number of significant figures as the least precise factor. So the statement saying it should match the most precise figure is incorrect. The other rules are standard significant-figure conventions.

Question 4

$ 16.8 \mathrm{g} $ of metal carbonate on thermal decomposition gave 8 g of metallic oxide and $ 8.8 \mathrm{g} $ of carbon dioxide. They follow:

Accepted Answer

Law of conservation of mass. The reactant mass is 16.8 g, and the products are 8.0 g oxide plus 8.8 g carbon dioxide, which also totals 16.8 g. Since mass is neither created nor destroyed, this illustrates the law of conservation of mass.

Question 5

Calculate the percent composition of carbon in $ C_{6} H_{12} O_{6} $

Accepted Answer

$ 35.3 \% $. The percent composition of an element is its mass in one mole of the compound divided by the compound’s molar mass, times 100. For glucose, carbon contributes 72 g out of a total 180 g, giving 35.3%.

Question 6

Which one of the followings has maximum number of atoms?

Accepted Answer

1 g of Li(s) [Atomic mass of Li = 7]. To find the maximum number of atoms, calculate the moles for each substance (mass/atomic mass) and multiply by Avogadro's number. Li has the smallest atomic mass, so 1 g of Li contains the most moles and hence the most atoms.

Question 7

In which case is number of molecules of water maximum?

Accepted Answer

18 mL of water. To determine the number of molecules, we calculate moles for each case. 18 mL of water corresponds to 1 mole (18 g), which contains the maximum number of molecules (6.022 × 10²³). Other options yield fewer moles.

Question 8

If Avogadro number NA is changed from 6.022 × 10²³ mol⁻¹ to 6.022 × 10²⁰ mol⁻¹ this would change:

Accepted Answer

the mass of one mole of carbon. If Avogadro's number changes, the mass of one mole of a substance (e.g., carbon) would change because the number of particles in a mole is directly tied to Avogadro's number. However, the ratios in equations or compounds remain unaffected.

Question 9

The number of moles of hydrogen molecule required to produce 20 moles of ammonia through Haber’s process is:

Accepted Answer

30. In the Haber process, the balanced equation is N2 + 3H2 → 2NH3. To produce 2 moles of ammonia, 3 moles of hydrogen are required. For 20 moles of ammonia, (20 × 3) / 2 = 30 moles of hydrogen are needed.

Question 10

A mixture of gases contains H2 and O2 gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the two gases in the mixture?

Accepted Answer

16:1. The molar mass of H2 is 2 g/mol, and that of O2 is 32 g/mol. For a 1:4 weight ratio, 1 g of H2 corresponds to 0.5 moles, and 4 g of O2 corresponds to 0.125 moles. Thus, the molar ratio is 0.5:0.125 = 4:1.

Question 11

When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g), each at S.T.P., the moles of HCl(g) formed is equal to:

Accepted Answer

1 mole of HCl(g). At STP, 1 mole of any gas occupies 22.4 L. The reaction H2 + Cl2 → 2HCl shows a 1:1 molar ratio between H2 and Cl2. Since 11.2 L of Cl2 (0.5 moles) is the limiting reagent, it reacts with 0.5 moles of H2 to form 1 mole of HCl.

Question 12

The number of atoms in 0.1 mol of a triatomic gas is: (NA = 6.02 × 10^23 mol^-1)

Accepted Answer

3.600 × 10^23. A triatomic gas contains 3 atoms per molecule. The total number of molecules in 0.1 mol is 0.1 × NA. Multiplying this by 3 gives the total number of atoms: 0.1 × 6.02 × 10^23 × 3 = 1.806 × 10^23.

Question 13

Which has the maximum number of molecules among the following?

Accepted Answer

8 g H2. To determine the number of molecules, calculate moles for each substance using their molar masses and multiply by Avogadro's number. H2 has the smallest molar mass (2 g/mol), so 8 g of H2 corresponds to the highest number of moles and thus the maximum number of molecules.

Question 14

What volume of oxygen gas (O2) measured at 0°C and 1 atm, is needed to burn completely 11.2 L of propane gas (C3H8) measured under the same conditions?

Accepted Answer

10L. The combustion reaction of propane is C3H8 + 5O2 → 3CO2 + 4H2O. From the stoichiometry, 1 mole of propane reacts with 5 moles of oxygen. Since gases at the same conditions have volumes proportional to moles, 11.2 L of propane will require 11.2 × 5 = 56 L of oxygen.

Question 15

Volume occupied by one molecule of water (density = 1 g cm^-3) is:

Accepted Answer

9.0 × 10^-23 cm^3. The molar mass of water is 18 g/mol, and its density is 1 g/cm³. Using Avogadro's number (6.022 × 10^23 molecules/mol), the volume of one molecule is calculated as (18 g / 1 g/cm³) ÷ 6.022 × 10^23, which equals 9.0 × 10^-23 cm³.

Question 16

1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reactant is left in excess and how much? (At. wt. Mg = 24; O = 16)

Accepted Answer

Mg, 0.16 g. The reaction is 2Mg + O2 → 2MgO. Moles of Mg = 1.0/24 = 0.0417, and moles of O2 = 0.56/32 = 0.0175. Mg requires 0.0175 × 2 = 0.035 moles, leaving 0.0417 - 0.035 = 0.0067 moles of Mg unreacted, which is 0.0067 × 24 = 0.16 g.

Question 17

An element, X has the following isotopic composition: 200X: 90%; 199X: 8.0%; 202X: 2.0%. The weighted average atomic mass of the naturally occurring element X is closest to:

Accepted Answer

200 amu. The weighted average atomic mass is calculated as (200 × 0.90) + (199 × 0.08) + (202 × 0.02) = 200 amu. Thus, the closest value is 200 amu.

Question 18

6.02 × 10^20 molecules of urea are present in 100 mL of its solution. The concentration of solution is:

Accepted Answer

0.001 M. The number of moles is obtained by dividing the given molecules by Avogadro's number. Molarity is then calculated using the solution volume.

Question 19

In an experiment it showed that 10 mL of 0.05 M solution of chloride required 10 mL of 0.1 M solution of AgNO3. Which of the following will be the formula of the chloride (X stands for the symbol of the element other than chlorine)?

Accepted Answer

XCl2. The reaction between chloride and AgNO3 is a precipitation reaction forming AgCl. The stoichiometry shows that 1 mole of chloride reacts with 1 mole of AgNO3. Since 10 mL of 0.05 M chloride reacts with 10 mL of 0.1 M AgNO3, the chloride must have 2 Cl atoms per formula unit, making the formula XCl2.

Question 20

10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be:

Accepted Answer

4 mol. The reaction is 2H2 + O2 → 2H2O. Moles of H2 = 10/2 = 5 mol, and moles of O2 = 64/32 = 2 mol. O2 is the limiting reagent, producing 2 × 2 = 4 mol of H2O.

NEET Chemistry: Some Basic Concepts of Chemistry questions with solutions

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