Q: Calculate the molar mass of the following substances.(a) Ethyne, C2H2(b) Sulphur molecule, S8(c) Phosphorus molecule, P4(Atomic mass of phosporus =31)(d) Hydrochloric acid, HCl(e) Nitric acid, HNO3

(a) Ethyne (C2H2) = 2xmass of atom +2xmass of H Atom =2x12+2x1=26 g/mol(b) Sulphur dioxide(S8)= 8xmass of sulphur atom =8x32=256g/mol(c) P4 = 4xMass of P atom =4x31 =124g/mol (d) HCL = 1x Mass of H atom+1xmass of Cl atom= 1+35.5 =36.5g/mol(e) HNO3 =1x mass of H atom +1x mass of N atom +3x Mass of O atom =1x1+1x14+3x16 = 63g/mol

Q: Calculate the number of molecules of sulphur(S8) present in 16 g of solid sulphur.

Moles of S8 = 25616=161 molNumber of molecules =16NA=166.023×1023=3.76×1022

Question 1

A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Accepted Answer

Boron and oxygen compound -> Boron + Oxygen0.24 g -> 0.096 g + 0.144 gPercentage composition, of the compoundFor boron:0.24 g→0.096 g100 g→?0.24100×0.096​=40%For oxygen:0.24 g→0.144 g of oxygen 100 g→?0.24100×0.144​=60%

Question 2

When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced? What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Accepted Answer

The reaction of burning of carbon in oxygen may be written as:It shows that 12 g of carbon bums in 32 g oxygen to form 44 g of carbon dioxide. Therefore 3 g of carbon reacts with 8 g of oxygen to form 11 g of carbon dioxide. It is given that 3.0 g of carbon is burnt with 8 g of oxygen to produce 11.0 g of CO2​ . Consequently 11.0 g of carbon dioxide will be formed when 3.0 g of C is burnt in 50 g of oxygen consuming 8 g of oxygen, leaving behind 50 €“ 8 = 42 g of O2​ . The answer governs the law of constant proportion.

Question 3

What are polyatomic ions? Give examples.

Accepted Answer

A polyatomic ion, also known as a molecular ion, is a charged chemical species(ion) composed of two or more atoms covalently bonded or of a metal complex that can be considered to be acting as a single unit.Eg: SO42−​, CO32−​.

Question 4

Write the chemical formula of the following.(a) Magnesium chloride(b) Calcium oxide(c) Copper nitrate(d) Aluminium chloride(e) Calcium carbonate

Accepted Answer

Chemical formula of the following is as follows:(a) Magnesium chloride = MgCl2​(b) Calcium oxide = CaO(c) Copper nitrate = Cu(NO3​)2​(d) Aluminium chloride = AlCl3​(e) Calcium carbonate = CaCO3​

Question 5

Give the names of the elements present in the following compounds.(a) Quick lime(b) Hydrogen bromide(c) Baking bromide(d) Potassium sulphate.

Accepted Answer

Compound Chemical FormulaElements present  Quick limeCaOCalcium, Oxygen Hydrogen bromide HBrHydrogen, Bromine Baking powderNaHCO3​Sodium, Hydrogen, Carbon, Oxygen  Potassium sulphateK2​SO4​ Potassium, Sulphur, Oxygen

Question 6

Calculate the molar mass of the following substances.(a) Ethyne, C2​H2​(b) Sulphur molecule, S8​(c) Phosphorus molecule, P4​(Atomic mass of phosporus =31)(d) Hydrochloric acid, HCl(e) Nitric acid, HNO3​

Accepted Answer

(a) Ethyne (C2​H2​) = 2xmass of atom +2xmass of H Atom =2x12+2x1=26 g/mol(b) Sulphur dioxide(S8​)= 8xmass of sulphur atom =8x32=256g/mol(c) P4​ = 4xMass of P atom =4x31 =124g/mol (d) HCL = 1x Mass of H atom+1xmass of Cl atom= 1+35.5 =36.5g/mol(e) HNO3​ =1x mass of H atom +1x mass of N atom +3x Mass of O atom =1x1+1x14+3x16 = 63g/mol

Question 7

What is the mass of:(a) 1 mole of nitrogen atoms?(b) 4 moles of aluminium atoms(Atomic mass of aluminium =27)?(c) 10 moles of sodium sulphite (Na2​SO3​)?

Accepted Answer

(a) 1 mole of nitrogen atoms = 1xmass of N = 1 x 14 = 14 g(b) 4 moles of aluminium atoms = 4x27 = 108 g(c) 10 moles of sodium sulphite( Na2​SO3​ )=10 x Mass of Na2​SO3​Mass of Na2​SO3​ =2xmass of N atom +1xmass of S atom +3xMass of O atom= 2x23+1x32+3x16=46+32+48= 126 g / mol10 moles of Na2​SO3​ = 10x126= 1260 g

Question 8

Convert into mole.(a) 12 g of oxygen gas(b) 20 g of water(c) 22 g of carbon dioxide.

Accepted Answer

(a) Molecular weight of oxygen in grams = 32 g of oxygen= one mole of oxygen gasThus, 12 g of oxygen=321​×12=0.375 moles(b) Molecular weight of water (H2​O) in grams= 2+ 16=18 g=one mole of waterThus, 20 g of water=181​×20=1.111 moles(c) Molecular weight of carbon dioxide (CO2​) in grams=12+2×16=44 g=one mole of carbon dioxideThus, 22 g of carbon dioxide =441​×22=0.5 moles

Question 9

What is the mass of : (a) 0.2 mole of oxygen atoms?  (b) 0.5 mole of water molecules?

Accepted Answer

(a) The mass of 1 mole of oxygen atoms = 16 gThus, the mass of 0.2 mole of oxygen atoms = 16 × 0.2  = 3.2 gb) The mass of 1 mole of water molecules = 18 gThus, the mass of 0.5 mole of water molecules = 18 × 0.5 = 9.0 g

Question 10

Calculate the number of molecules of sulphur(S8​) present in 16 g of solid sulphur.

Accepted Answer

Moles of S8​ = 25616​=161​ molNumber of molecules =16NA​​=166.023×1023​=3.76×1022

Question 11

Calculate the number of aluminum ions present in 0.051g of aluminum oxide.

Accepted Answer

Molar mass of aluminium oxide Al2​O3​ = (2 x 27) + (3 x 16)= 54 + 48 = 102 g.∴102 g of Al2​O3​ contains =2×6.022×1023 aluminium ions ∴0.051 g of Al2​O3​ contains =1022×6.022×1023​×0.051=10212.044×1023×0.051​=1020.614×1023​=0.006022×1023=6.022×1020Al3+ ions

Question 12

What is the chemical formula of potassium nitride?

Accepted Answer

Valency of potassium is 1 and that of nitride ion is 3. Thus, by criss-cross method, the formula of potassium nitride comes out to be K3​N.

Question 13

Sodium carbonate reacts with ethanoic acid to form sodium ethanoate, carbon dioxide and water. In an experiment, 9.7 g of sodium carbonate reacted with 15.3 g of ethanoic acid to form 5.9 g of sodium ethanoate, 2.4 g of carbon dioxide and x g of water. The value of x is ___.

Accepted Answer

The reaction taking place is:Sodium carbonateNa2​CO3​​+Ethanoic acidCH3​COOH​→Sodium ethanoateCH3​COONa​+Carbon dioxideCO2​​+WaterH2​O​According to Law of conservation of mass, total mass of reactants is equal to the total mass of products.9.7+15.3=5.9+2.4+xx=16.7 g

Question 14

Which of the following is a trivalent ion?

Accepted Answer

The chemical formulae of nitride, sulphide, sulphite and oxide are N−3,S−2,SO3−2​ and O−2 respectively.
An ion having magnitude of charge as 3 is considered as the trivalent ion. Thus, nitride is a trivalent ion.

Question 15

A 0.96 g sample of compound of oxygen and boron was found by analysis to contain 0.72 g of oxygen. Calculate the percentage composition of boron in the compound by mass.

Accepted Answer

Mass of oxygen in compound =0.72 gAnd,  Mass of compound =0.96 gMass of boron in compound=0.96−0.72 or 0.24 gPercentage of boron=Mass of compoundMass of boron​×100Percentage of boron=0.960.24​×100 or 25%

Question 16

10 g calcium carbonate on heating gives 4.4 g calcium oxide. What is the percentage purity of the sample?

Accepted Answer

The reaction taking place is:CaCO3​Heat​CaO+CO2​  100 g             56 g       44 g56 g calcium oxide is produced 100 g calcium carbonate.4.4 g calcium oxide will be produced by 7.86 g calcium carbonate.Percentage purity =Mass of impure sampleMass of pure sample​×100Percentage purity=107.86​×100=78.6%

Question 17

The mass of 1 molecule of water is :

Accepted Answer

We know that 1 mole of any particle = Relative mass of that particle in grams =6.022×1023 number of that particles Since, water is a molecule and has molecular mass of 18u.So, 1 mole of water molecule  =18g=6.022×1023 number of water molecules.⇒6.022×1023 number of water molecules =18g∴1 molecule of water =6.022×102318​g=2.989×10−23g

Question 18

The number of gram molecules of oxygen in 6.02×1024 CO molecules is -

Accepted Answer

We know that 1 mole of any particle = Relative mass of that particle in grams =6.022×1023 number of that particles 1 mole of CO=6.022×1023 molecules of CO.Since, in one molecule of CO, one oxygen atom is present.So, 6.022×1023 molecules of CO contains 6.022×1023 atoms of O.⇒6.02×1024 molecules of CO  contains 6.02×1024  atoms of O.∴ number of gram atoms of O=6.02×10236.02×1024​=10g atom As oxygen is a diatomic molecule. So, the number of gram molecules of oxygen =2atomspermolecule10g atom​=5 g molecules.

Question 19

State whether the following statement is  true or false. Formula for potassium biphosphate is KHPO4​.

Accepted Answer

The potassium ion is represented as K+ and biphosphate ion is represented as HPO42−​.By criss-cross method, formula is K2​HPO4​.Hence, the given statement is false.

Question 20

Question 1:
Which of the following correctly represents 360g of water?
(i) 2 moles of H2O
(ii) 20 moles of water
(iii) 6.022 × 1023 molecules of water
(iv) 1.2044 ×1025 molecules of water

Accepted Answer

Answer:
(i) 2 moles of water = 18×2 = 36 grams of water(ii) 20 moles of water = 18×20 = 360 grams of water(iii) 6.022×1023 molecules of water = 1 mole of water = 18 grams of water(iv) 1.2044 × 1025 molecules of water = 6.022×10231.2044×1025​=20 moles of water = 360 grams of water
Hence, the correct answer is option d.

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