Q: Calculate the number of atoms in(i) half a mole of nitrogen gas.(ii) 5.6 litres of helium gas at S.T.P.

(i) One mole of N2=6.022×1023 molecules ⇒0.5 moles of N2=26.022×1023=3.015×1023 molecules.Since N2 is diatomic, the number of atoms of N2= 3.015×1023×2=6.022×1023 atoms(ii) From Avogadro's hypothesis, 22.4 litres of gas at S.T.P are occupied by one mole.5.6 litres of gas at S.T.P are occupied by 22.45.6=0.25 molesNo. of atoms in 0.25moles=6.022×1023×0.25=1.51 ×1023 atoms.

Q: In a reaction where an electron transfer takes place from one species (atom, molecule, ion etc.) to another, the species which gives electrons is termed "reducing agent" and gets oxidized. The species which accepts electrons is called an "oxidizing agent" and gets reduced in the reaction. Each species may be assigned a "so-called" oxidation number, according to certain rules. Increase in this number is oxidation and decrease is reduction.Thus,Oxidation number change: negative → zero → positive valueThis is oxidation. The changes indicated by reversed arrows is reduction. Equivalent weight of the species = Change in oxidation number Mol.weight 1. When KMnO4 reacts with FeSO4 in dilute H2SO4 medium it forms MnSO4. What is the decrease in oxidation number for Mn in this reaction?

1. MnO4−changes to Mn2+2. Since the change in ON is by five units equivalent weight =31.63. Equivalent weight. of FeSO4.(NH4)2SO4.6H2O in the reaction is 392

Q: An inorganic compound, on analysis, gave the following weight percent composition. Mg=9.86;S=13.00; O=71.41, the rest being hydrogen. Assuming that all the hydrogen is in the form of water, calculate its empirical formula. If its molecular weight is 246.5, what is the molecular formula of the compound?(Atomic mass of Mg=24.3;S=32.06;O=16.00;H=1 )

Weight of carbon =1310×93=71.5 g;1271.5=6 g atom. Weight of hydrogen =131×93=7.15 g;17.15=7 g atom. Weight of nitrogen =132×93=14.3 g;1414.3=1 g atom. ∴ Molecular formula is C6H7 N(C6H5NH2)

Q: A compound contains carbon, hydrogen and nitrogen in the weight ratio 10:1:2. If the molecular weight of the compound is 93 g mol−1, find its molecular formula.

SolutionWeight of carbon =1310×93=71.5 g;1271.5=6 g atom.Weight of hydrogen =131×93=7.15 g;17.15=7 g atom.Weight of nitrogen =132×93=14.3 g;1414.3=1 g atom.∴ Molecular formula is C6H7 N(C6H5NH2)

Q: The ratio of weights of mercurous chloride to mercury is 2.354. Find the atomic mass of mercury. (This example illustrates the importance of the use of correct formula for a compound.)

Mercurous chloride is Hg2Cl2. If x is the atomic mass of Hg, then x2x+(2×35.5)=2.354Solving for x, we get atomic mass of Hg as 200.6If we use the formula HgCl, the answer becomes different.

Q: Calculate the weight of hydrochloric acid required to react with 5 g of calcium carbonate completely. How much of CO2 is formed under the given conditions?

(i) Oxidation number of Mn in K2MnO4 is +2+x−8= 0,x=+6(ii) Oxidation number of Cr in K2Cr2O7 is +2+2x− 14=0,x=+6(iii) Oxidation number of S in Na2 S2O3 is +2+2x− 6=0,x=+2In fact, the two sulphur atoms are in +6 and -2 oxidation states and the average works out to be +2 .(iv) Oxidation number of C in CH3COOH(C2H4O2) is 2x+4−4=0,x=0(v). Oxidation number of N in HN3 is −31.(vi) Oxidation number of Cr in CrO5 is +6 . (since, 4×−1+ x+1×−2=0⇒x=+6)

Q: Calculate the weight of hydrochloric acid required to react with 5 g of calcium carbonate completely. How much of CO2 is formed under the given conditions?

Balanced equation:CaCO3+2HCl→CaCl2+H2O+CO21 mole of CaCO3 react with 2 moles of HCl for complete reactioni.e., 100 g of CaCO3 react with 2×36.5=73.0 g of HCl 5 g of CaCO3 react with 1005×73.0=3.65 g of HClMass of CO2 formed:100 g of CaCO3 on reaction with HCl gives 44 g of CO2. 5 g of CaCO3 on complete reaction with HCl gives1005×44=2.20 gof CO2.

Question 1

Calculate the number of atoms in(i) half a mole of nitrogen gas.(ii) 5.6 litres of helium gas at S.T.P.

Accepted Answer

(i) One mole of N2​=6.022×1023 molecules ⇒0.5 moles of N2​=26.022×1023​=3.015×1023 molecules.Since N2​ is diatomic, the number of atoms of N2​= 3.015×1023×2=6.022×1023 atoms(ii) From Avogadro's hypothesis, 22.4 litres of gas at S.T.P are occupied by one mole.5.6 litres of gas at S.T.P are occupied by 22.45.6​=0.25 molesNo. of atoms in 0.25moles=6.022×1023×0.25=1.51 ×1023 atoms.

Question 2

Statement 1When white phosphorus is heated with NaOH both PH3​ and NaH2​PO2​ are obtained as products. andStatement 2In such reactions eg: Cl2​+NaOH or S+NaOH etc. the same element suffers both an increase and a decrease of its oxidation state/number.

Accepted Answer

Statement (1) and (2) are correct. Statement (2) is the correct explanation to statement (1)
When white phosphorus is heated with NaOH both PH3​ and NaH2​PO2​ are obtained as products. In such reactions eg: Cl2​+NaOH or S+NaOH etc. the same element suffers both an increase and a decrease of its oxidation state/number.

Question 3

In a reaction where an electron transfer takes place from one species (atom, molecule, ion etc.) to another, the species which gives electrons is termed "reducing agent" and gets oxidized. The species which accepts electrons is called an "oxidizing agent" and gets reduced in the reaction. Each species may be assigned a "so-called" oxidation number, according to certain rules. Increase in this number is oxidation and decrease is reduction.Thus,Oxidation number change: negative → zero → positive valueThis is oxidation. The changes indicated by reversed arrows is reduction. Equivalent weight of the species = Change in oxidation number  Mol.weight ​1. When KMnO4​ reacts with FeSO4​ in dilute H2​SO4​ medium it forms MnSO4​. What is the decrease in oxidation number for Mn in this reaction?

Accepted Answer

1. MnO4−​changes to Mn2+2. Since the change in ON is by five units equivalent weight =31.63. Equivalent weight. of FeSO4​.(NH4​)2​SO4​.6H2​O in the reaction is 392

Question 4

Match the reactions on the left with the types of reactions on the right.

Accepted Answer

(a)→(p),(r),(s)(b) → (p)(c) → (p), (s)(d)→(p),(q)

Question 5

A molecule of a diatomic compound weighs 4.98×10−23 g. What is its molecular weight? How many molecules and atoms are present in 30 kg of this compound?

Accepted Answer

Molecular weight of this compound =4.98×10−23×6.022 ×1023=30Number of moles in 30 kg of this compound =3030×103​=1000No. of molecules in 30 kg=1000×6.022×1023 =6.022×1026No. of atoms in 30 kg=1000×6.022×1023×2 =1.2046×1027

Question 6

An inorganic compound, on analysis, gave the following weight percent composition. Mg=9.86;S=13.00; O=71.41, the rest being hydrogen. Assuming that all the hydrogen is in the form of water, calculate its empirical formula. If its molecular weight is 246.5, what is the molecular formula of the compound?(Atomic mass of Mg=24.3;S=32.06;O=16.00;H=1 )

Accepted Answer

Weight of carbon =1310​×93=71.5 g;1271.5​=6 g atom. Weight of hydrogen =131​×93=7.15 g;17.15​=7 g atom. Weight of nitrogen =132​×93=14.3 g;1414.3​=1 g atom. ∴ Molecular formula is C6​H7​ N(C6​H5​NH2​)

Question 7

A compound contains carbon, hydrogen and nitrogen in the weight ratio 10:1:2. If the molecular weight of the compound is 93 g mol−1, find its molecular formula.

Accepted Answer

SolutionWeight of carbon =1310​×93=71.5 g;1271.5​=6 g atom.Weight of hydrogen =131​×93=7.15 g;17.15​=7 g atom.Weight of nitrogen =132​×93=14.3 g;1414.3​=1 g atom.∴ Molecular formula is C6​H7​ N(C6​H5​NH2​)

Question 8

The ratio of weights of mercurous chloride to mercury is 2.354. Find the atomic mass of mercury. (This example illustrates the importance of the use of correct formula for a compound.)

Accepted Answer

Mercurous chloride is Hg2​Cl2​. If x is the atomic mass of Hg, then x2x+(2×35.5)​=2.354Solving for x, we get atomic mass of Hg as 200.6If we use the formula HgCl, the answer becomes different.

Question 9

Calculate the weight of hydrochloric acid required to react with 5 g of calcium carbonate completely. How much of CO2​ is formed under the given conditions?

Accepted Answer

(i) Oxidation number of Mn in K2​MnO4​ is +2+x−8= 0,x=+6(ii) Oxidation number of Cr in K2​Cr2​O7​ is +2+2x− 14=0,x=+6(iii) Oxidation number of S in Na2​ S2​O3​ is +2+2x− 6=0,x=+2In fact, the two sulphur atoms are in +6 and -2 oxidation states and the average works out to be +2 .(iv) Oxidation number of C in CH3​COOH(C2​H4​O2​) is 2x+4−4=0,x=0(v). Oxidation number of N in HN3​ is −31​.(vi) Oxidation number of Cr in CrO5​ is +6 . (since, 4×−1+ x+1×−2=0⇒x=+6)

Question 10

Calculate the weight of hydrochloric acid required to react with 5 g of calcium carbonate completely. How much of CO2​ is formed under the given conditions?

Accepted Answer

Balanced equation:CaCO3​+2HCl→CaCl2​+H2​O+CO2​1 mole of CaCO3​ react with 2 moles of HCl for complete reactioni.e., 100 g of CaCO3​ react with 2×36.5=73.0 g of HCl 5 g of CaCO3​ react with 1005×73.0​=3.65 g of HClMass of CO2​ formed:100 g of CaCO3​ on reaction with HCl gives 44 g of CO2​. 5 g of CaCO3​ on complete reaction with HCl gives1005×44​=2.20 gof CO2​.

Question 11

20.0 g of silver nitrate in aqueous solution react with 6.0 g sodium chloride to form silver chloride precipitate. Calculate the amount of silver chloride formed and specifywhich among the reactants, is an (Atomic mass of Ag= 108,Cl=35.5,Na=23 ).(i) excess reactant(ii) limiting reactant.

Accepted Answer

The reaction is AgNO3​+NaCl→AgCl↓+NaNO3​1 mole of AgNO3​ reacts with one mole of NaCl to give one mole of AgCl.i.e., 170 g of AgNO3​ reacts with 58.5 g of NaCl to give 143.5 g of AgCl.6.00 g of NaCl react completely with 58.56×170​=17.44 g of AgNO3​.Since 20.00−17.44=2.56 g of AgNO3​ remains in solution, the excess reactant is AgNO3​ and as NaCl limits the amount of AgCl formed, it is the limiting reactant. Amount of AgCl formed =58.56×143.5​=14.72 g

Question 12

20.0 g of silver nitrate in aqueous solution react with 6.0 g sodium chloride to form silver chloride precipitate. Calculate the amount of silver chloride formed and specifywhich among the reactants, is an (Atomic mass of Ag= 108,Cl=35.5,Na=23 ).(i) excess reactant(ii) limiting reactant.

Accepted Answer

The reaction is AgNO3​+NaCl→AgCl↓+NaNO3​1 mole of AgNO3​ reacts with one mole of NaCl to give one mole of AgCl.i.e., 170 g of AgNO3​ reacts with 58.5 g of NaCl to give 143.5 g of AgCl.6.00 g of NaCl react completely with 58.56×170​=17.44 g of AgNO3​.Since 20.00−17.44=2.56 g of AgNO3​ remains in solution, the excess reactant is AgNO3​ and as NaCl limits the amount of AgCl formed, it is the limiting reactant.Amount of AgCl formed =58.56×143.5​=14.72 g

Question 13

Calculate the oxidation number of(i) Mn in K2​MnO4​(ii) Cr in K2​Cr2​O7​(iii) S in Na2​ S2​O3​(iv) C in CH3​COOH(v) N in HN3​(vi) Cr in CrO5​In fact, the two sulphur atoms are in +6 and −2 oxidation states and the average works out to be +2.(iv) Oxidation number of C in CH3​COOH(C2​H4​O2​) is 2x+4−4=0,x=0(v). Oxidation number of N in HN3​ is −31​.

Accepted Answer

(i) Oxidation number of Mn in K2​MnO4​ is +2+x−8= 0,x=+6(ii) Oxidation number of Cr in K2​Cr2​O7​ is +2+2x− 14=0,x=+6(iii) Oxidation number of S in Na2​ S2​O3​ is +2+2x− 6=0,x=+2In fact, the two sulphur atoms are in +6 and -2 oxidation states and the average works out to be +2 .(iv) Oxidation number of C in CH3​COOH(C2​H4​O2​) is 2x+4−4=0,x=0(v). Oxidation number of N in HN3​ is −31​.(vi) Oxidation number of Cr in CrO5​ is +6 . (since, 4×−1+ x+1×−2=0⇒x=+6)

Question 14

Calculate the oxidation number of(i) Mn in K2​MnO4​(ii) Cr in K2​Cr2​O7​(iii) S in Na2​ S2​O3​(iv) C in CH3​COOH(v) N in HN3​(vi) Cr in CrO5​In fact, the two sulphur atoms are in +6 and −2 oxidation states and the average works out to be +2.(iv) Oxidation number of C in CH3​COOH(C2​H4​O2​) is 2x+4−4=0,x=0(v). Oxidation number of N in HN3​ is −31​.

Accepted Answer

(i) Oxidation number of Mn in K2​MnO4​ is +2+x−8= 0,x=+6
(ii) Oxidation number of Cr in K2​Cr2​O7​ is +2+2x− 14=0,x=+6
(iii) Oxidation number of S in Na2​ S2​O3​ is +2+2x− 6=0,x=+2

In fact, the two sulphur atoms are in +6 and -2 oxidation states and the average works out to be +2 .
(iv) Oxidation number of C in CH3​COOH(C2​H4​O2​) is 2x+4−4=0,x=0
(v). Oxidation number of N in HN3​ is −31​.
(vi) Oxidation number of Cr in CrO5​ is +6 . (since, 4×−1+ x+1×−2=0⇒x=+6)

Question 15

Balance the equation: H2​SO4​+HI→H2​ S+H2​O+I2​

Accepted Answer

For making the increase and decrease in the oxidation numbers the same, use the coefficients 8 for HI i.e., H2​SO4​+8HI→H2​ S+H2​O+I2​ Balancing the other atomsH2​SO4​+8HI→H2​ S+4I2​+4H2​O

Question 16

Balance the equationK2​Cr2​O7​+H2​SO4​+FeSO4​→K2​SO4​+Cr2​(SO4​)3​+Fe2​(SO4​)3​+H2​O

Accepted Answer

SolutionUsing the coefficient 6 for FeSO4​K2​Cr2​O7​+H2​SO4​+6FeSO4​→K2​SO4​+Cr2​(SO4​)3​+Fe2​(SO4​)3​+H2​O​Balancing all the atomsK2​Cr2​O7​+6FeSO4​+7H2​SO4​→K2​SO4​+Cr2​(SO4​)3​+3Fe2​(SO4​)3​+7H2​O​

Question 17

Balance the equationK2​Cr2​O7​+H2​SO4​+FeSO4​→K2​SO4​+Cr2​(SO4​)3​+Fe2​(SO4​)3​+H2​O

Accepted Answer

SolutionUsing the coefficient 6 for FeSO4​K2​Cr2​O7​+H2​SO4​+6FeSO4​→K2​SO4​+Cr2​(SO4​)3​+Fe2​(SO4​)3​+H2​O​Balancing all the atomsK2​Cr2​O7​+6FeSO4​+7H2​SO4​→K2​SO4​+Cr2​(SO4​)3​+3Fe2​(SO4​)3​+7H2​O​>

Question 18

Represent the disproportionation reaction of chlorine in alkaline medium.

Accepted Answer

Cl2​ disproportionates to (+1) and (−1) states.

Filters

Filters

book

topic

difficulty

Solutions for IIT-JEE Super Course in Chemistry - Physical Chemistry by Trishna Knowledge Systems

Found the solution, but did not understand the concept?

Ask my first question