Question 1

What is the number of molecules in 2.125 g of ammonia?

Accepted Answer

Molar mass of ammonia (NH_3) = 14 + 3 = 17 g/mol Number of moles of ammonia = 2.125 17 = 0.125 mol Number of molecules = 0.125 6.022 10^ 23 = 7.527 10^ 22 Answer: 7.527 10^ 22

Question 2

What is the SI unit of luminous intensity ?

Accepted Answer

Luminous intensity is one of the seven fundamental base quantities in the International System of Units (SI). The SI unit of luminous intensity is the candela (cd). Ampere is the SI unit of electric current, Coulomb is the SI unit of electric charge, and Ohm is the SI unit of electrical resistance. Answer: Candela

Question 3

What is average atomic mass of chlorine if its two isotopes Cl ^ 35 and Cl ^ 37 exists in relative abundance of 75\% and 25\% respectively ?

Accepted Answer

Average atomic mass is calculated using the formula: Average atomic mass = m_1 p_1 + m_2 p_2 100 where m_1 and m_2 are the atomic masses of the isotopes, and p_1 and p_2 are their respective percentage abundances. Substituting the given values: Average atomic mass = 35 75 + 37 25 100 Average atomic mass = 2625 + 925 100 Average atomic mass = 3550 100 = 35.5 Answer: 35.5

Question 4

Chlorine has two isotopes Cl ^ 37 and Cl ^ 35 with percentage abundance of 25\% and 75\% respectively. Find average atomic mass of chlorine ?

Accepted Answer

Average atomic mass is calculated using the formula: Average atomic mass = ( Isotopic mass Percentage abundance ) 100 Substituting the given values: Average atomic mass = 37 25 + 35 75 100 Average atomic mass = 925 + 2625 100 Average atomic mass = 3550 100 = 35.5 Answer: 35.5

Question 5

What quantity of H _2 O in gram is present in 0.25 mol of it?

Accepted Answer

Molar mass of H _2 O = 2 1 + 16 = 18 g/mol Mass of H _2 O = moles molar mass Mass = 0.25 18 = 4.5 g Answer: 4.5 g

Question 6

What is the percentage atom economy when formula weight of product is 65 u and the sum of formula weight of reactant is 130 u?

Accepted Answer

The percentage atom economy is calculated using the formula: Percentage atom economy = Formula weight of desired product Sum of formula weights of all reactants 100 Substituting the given values: Percentage atom economy = 65 130 100 = 50\% Answer: 50\%

Question 7

Find the number of molecules present in 70 g dinitrogen.

Accepted Answer

Molar mass of dinitrogen (N_2) is 28 g mol ^ -1 Number of moles of N_2 = 70 28 = 2.5 mol Number of molecules = n N_A Number of molecules = 2.5 6.022 10^ 23 = 1.5055 10^ 24 Answer: 1.5055 10^ 24

Question 8

How many moles of potassium chlorate are heated to produce 22.4\ lit of oxygen at STP ?

Accepted Answer

The balanced chemical equation for the thermal decomposition of potassium chlorate is: 2KClO_3 2KCl + 3O_2 At STP, 22.4\ lit of oxygen gas corresponds to 1\ mole of O_2. From the stoichiometry of the reaction, 3\ moles of O_2 are produced by heating 2\ moles of KClO_3. Therefore, the number of moles of KClO_3 required to produce 1\ mole of O_2 is 2 3 \ mole . Answer: 2 3 \ mole

Question 9

How many moles of potassium chlorate are heated to produce 11 2 L oxygen at STP ?

Accepted Answer

The balanced chemical equation for the decomposition of potassium chlorate is: 2KClO_ 3 2KCl + 3O_ 2 At STP, the volume of 1 mole of an ideal gas is 22 4 L. Number of moles of O_ 2 produced is: n(O_ 2 ) = 11 2 22 4 = 1 2 mole From the stoichiometry of the reaction, 3 moles of O_ 2 are produced from 2 moles of KClO_ 3 . Therefore, the number of moles of KClO_ 3 required to produce 1 2 mole of O_ 2 is: n(KClO_ 3 ) = 2 3 n(O_ 2 ) = 2 3 1 2 = 1 3 mole Answer: 1 3 mole

Question 10

Calculate the number of molecules present in 2.24 dm ^3 of carbon dioxide at STP ?

Accepted Answer

Volume of CO _2 at STP = 2.24 dm ^3 = 2.24 L Molar volume of an ideal gas at STP = 22.4 L Number of moles of CO _2 = 2.24 22.4 = 0.1 mol Number of molecules = Number of moles N_A Number of molecules = 0.1 6.022 10^ 23 = 6.022 10^ 22 Answer: 6.022 10^ 22

Question 11

The ratio of mass of 'x' atoms of an element to mass of carbon atoms is 9:1. Find the mass of 1 mol of x if molar mass of carbon is 12 g/mol.

Accepted Answer

Let the mass of one atom of element 'x' be m_x and the mass of one carbon atom be m_C. Given that the ratio of the mass of an atom of element 'x' to the mass of a carbon atom is 9:1, we have: m_x m_C = 9 1 The ratio of their molar masses is equal to the ratio of their atomic masses: M_x M_C = 9 Given the molar mass of carbon, M_C = 12\ g/mol , the molar mass of element 'x' is: M_x = 9 12 = 108\ g/mol Therefore, the mass of 1\ mol of element 'x' is 108\ g .

Question 12

Calculate the amount of aluminium present in Al _2( SO _4)_3 if the compound contains 4 g of sulfur. [Molar mass of sulfur = 32 g/mol , Molar mass of Al =27 g/mol ]

Accepted Answer

In one mole of Al _2( SO _4)_3, there are 2 moles of Al and 3 moles of S. Mass of Al in 1 mole of compound = 2 27 = 54 g Mass of S in 1 mole of compound = 3 32 = 96 g The ratio of the mass of Al to the mass of S in the compound is 54 96 = 9 16 . Given the mass of S is 4 g, the mass of Al is calculated as: Mass of Al = 4 9 16 = 9 4 = 2.25 g Answer: 2.25 g

Question 13

What is number of hydrogen molecules needed to synthesize 3.4 g of ammonia by reaction with nitrogen?

Accepted Answer

The balanced chemical equation for the synthesis of ammonia is: N_2 + 3H_2 2NH_3 Molar mass of NH_3 = 14 + 3 1 = 17 g/mol Moles of NH_3 to be synthesized = 3.4 17 = 0.2 mol From the balanced equation, 2 moles of NH_3 require 3 moles of H_2. Moles of H_2 required = 3 2 0.2 = 0.3 mol Number of H_2 molecules = 0.3 6.022 10^ 23 = 1.8066 10^ 23 Answer: 1.8066 10^ 23

Question 14

Find the mass of sodium carbonate in grams to form 100 ml 0.1 M solution in water? ( Molar mass of Na _2 CO _3 = 106 g/mol)

Accepted Answer

Molarity is given by the formula: M = n V (in L) Substituting the given values: 0.1 = n 0.1 n = 0.01 moles Mass of Na _2 CO _3 is calculated as: w = n Molar mass w = 0.01 106 = 1.06 g Answer: 1.06 g

Question 15

What is the approximate mass of the precipitate formed when 50 mL of 16.9\% solution of AgNO _3 is mixed with 50 mL of 7.45\% KCl solution? (Molar mass of AgNO _3 = 169 g/mol, KCl = 74.5 g/mol, AgCl =

Accepted Answer

Mass of AgNO _3 in 50 mL of 16.9\% (w/v) solution = 16.9 100 50 = 8.45 g Moles of AgNO _3 = 8.45 169 = 0.05 mol Mass of KCl in 50 mL of 7.45\% (w/v) solution = 7.45 100 50 = 3.725 g Moles of KCl = 3.725 74.5 = 0.05 mol The balanced chemical equation is: AgNO _3 + KCl AgCl + KNO _3 Since 1 mole of AgNO _3 reacts with 1 mole of KCl to give 1 mole of AgCl, 0.05 mol of AgNO _3 will react completely with 0.05 mol of KCl to form 0.05 mol of AgCl. Mass of AgCl precipitate = 0.05 143.3 = 7.165 g 7 g Ans

Question 16

1 g of H _2 contains the same number of molecules as in

Accepted Answer

Number of moles of H _2 = 1 2 = 0.5 mol. Number of moles of N _2 in 14 g = 14 28 = 0.5 mol. Number of moles of H _2 O in 18 g = 18 18 = 1 mol. Number of moles of CO in 16 g = 16 28 = 4 7 mol. Number of moles of N _2 in 28 g = 28 28 = 1 mol. Since the number of moles of 1 g of H _2 and 14 g of N _2 are equal, they contain the same number of molecules. Answer: 14 g of N _2

Question 17

Which of the following contains maximum number of molecules?

Accepted Answer

The number of molecules is directly proportional to the number of moles. Number of moles of H _2 in 0.4 g = 0.4 2 = 0.2 mol Number of moles of O _2 in 3.2 g = 3.2 32 = 0.1 mol Number of moles of H _2 in 16.8 dm ^3 at STP = 16.8 22.4 = 0.75 mol Number of moles of O _2 in 11.2 dm ^3 at STP = 11.2 22.4 = 0.5 mol Since 0.75 mol is the maximum among the given options, 16.8 dm ^3 of H _2 gas at STP contains the maximum number of molecules. Answer: 16.8 dm ^3 of H _2 gas at STP

Question 18

Calculate the volume of 99 g of CO _2 at STP ?

Accepted Answer

Molar mass of CO _2 = 12 + 2 16 = 44 g/mol Number of moles of CO _2 = 99 44 = 2.25 moles Volume of 1 mole of a gas at STP = 22.4 L Volume of 2.25 moles of CO _2 = 2.25 22.4 = 50.4 L Answer: 50.4 L

Question 19

Calculate the mass of 100 molecules of oxygen in amu and in grams.

Accepted Answer

Mass of one oxygen molecule (O_2) = 2 16 u = 32 u. Mass of 100 oxygen molecules = 100 32 u = 3200 u. Since 1 u = 1.6605 10^ -24 g, the mass in grams is given by: Mass = 3200 1.6605 10^ -24 g = 5.314 10^ -21 g. Answer: 3200 u and 5.314 10^ -21 g

Some Basic Concepts of Chemistry — Previous Year Questions & Solutions