Q: Question 15Water decomposes to O2 and H2 under suitable conditions as represented by the equation below:2H2O ⟶ 2H2 + O2(a) If 2500 cm3 of H2 is produced, what volume of O2 is liberated at the same time and under the same conditions of temperature and pressure?(b) The 2500 cm3 of H2 is subjected to 2122\dfrac{1}{2}221 times increase in pressure (temp. remaining constant). What volume will H2 now occupy?(c) Taking the value of H2 calculated in 5(b), what changes must be made in Kelvin (absolute) temperature to return the volume to 2500 cm3 pressure remaining constant.

2H2O⟶2H2+O2 Vol.2 Vol.1 Vol.\begin{matrix} 2\text{H}_2\text{O} & \longrightarrow & 2\text{H}_2 & + &\text{O} \\ 2\text{ Vol.} & & 2\text{ Vol.}& & 1\text{ Vol.} \end{matrix}2H2O2 Vol.⟶2H22 Vol.+O1 Vol.2 Vol. of water gives 2 Vol. of H2 and 1 Vol. of O2∴ If 2500 cm3 of H2 is produced, volume of O2 produced = 25002\dfrac{2500}{2}22500 = 1250 cm3(b) V1 = 2500 cm3P1 = 1 atm = 760 mmT1 = TT2 = TP2 = [760 x 2 12\dfrac{1}{2}21] + [760] = 760 [52\dfrac{5}{2}25 + 1] = 760 x 72\dfrac{7}{2}27 = 2660 mmV2 = ?Using formula:P1V1T1\dfrac{\text{P}_1\text{V}_1}{\text{T}_1}T1P1V1 = P2V2T2\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}T2P2V2760×2500T\dfrac{760 \times 2500}{\text{T}}T760×2500 = 2660×V2T2660 \times \dfrac{\text{V}_2}{\text{T}}2660×TV2V2 = 760×25002660\dfrac{760 \times 2500 }{2660}2660760×2500 = 50007\dfrac{5000}{7}75000(c) V1 = 50007\dfrac{5000}{7}75000 = 714.29 cm3P1 = P2 = PT1 = TV2 = 2500 cm3T2 = ?Using formula:P1V1T1\dfrac{\text{P}_1\text{V}_1}{\text{T}_1}T1P1V1 = P2V2T2\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}T2P2V2P×714.29T\dfrac{\text{P} \times 714.29}{\text{T}}TP×714.29 = P×2500T2\dfrac{\text{P} \times 2500}{\text{T}_2}T2P×2500T2 = 2500714.29\dfrac{2500 }{714.29}714.292500 x TT2 = 3.5 x THence, T2 = 3.5 times T or temperature should be increased by 3.5 times

Question 1

Question 6What volume of oxygen at STP is required to affect the combustion of 11 litres of ethylene [C2H4] at 273°C and 380 mm of Hg pressure?C2H4 + 3O2 ⟶ 2CO2 + 2H2O

Accepted Answer

C2H4+3O2⟶2CO2+2H2O1 vol.:3 vol.11 lit:33 lit\begin{matrix} \text{C}_2\text{H}_4 & + & 3\text{O}_2 & \longrightarrow & 2\text{CO}_2 + 2\text{H}_2\text{O} \ 1 \text{ vol.} & : & 3 \text{ vol.} \ 11 \text{ lit} & : & 33 \text{ lit} \end{matrix}C2​H4​1 vol.11 lit​+::​3O2​3 vol.33 lit​⟶2CO2​+2H2​OSTPGiven ValuesP1 = 760 mm of HgP2 = 380 mm of HgV1 = x litV2 = 33 litT1 = 273 KT2 = 273 + 273 K = 546 KUsing the gas equation,P1V1T1=P2V2T2\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}T1​P1​V1​​=T2​P2​V2​​Substituting the values we get,760×x273=380×33546x=380×33×273546×760x=3,423,420414,960x=8.25 lit\dfrac{760 \times x}{273} = \dfrac{380 \times 33}{546} \[0.5em] x = \dfrac{380 \times 33 \times 273}{546 \times 760 } \[0.5em] x = \dfrac{3,423,420}{414,960} \ \[0.5em] x = 8.25 \text{ lit}273760×x​=546380×33​x=546×760380×33×273​x=414,9603,423,420​x=8.25 litHence, volume of oxygen required = 8.25 lit.

Question 2

Question 13What volume of air (containing 20% O2 by volume) will be required to burn completely 10 cm3 each of methane and acetylene?CH4 + 2O2 ⟶ CO2 + 2H2O2C2H2 + 5O2 ⟶ 4CO2 + 2H2O

Accepted Answer

CH4+2O2⟶CO2+2H2O1 vol.:2 vol.⟶1 vol.2C2H2+5O2⟶4CO2+2H2O2 vol.:5 vol.⟶4 vol.\begin{matrix} \text{CH}_4 & + & 2\text{O}_2 & \longrightarrow & \text{CO}_2 & + & 2\text{H}_2\text{O} \ 1 \text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 1\text{ vol.} \ 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} \ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} \ \end{matrix}CH4​1 vol.2C2​H2​2 vol.​+:+:​2O2​2 vol.5O2​5 vol.​⟶⟶⟶⟶​CO2​1 vol.4CO2​4 vol.​++​2H2​O2H2​O​[By Gay Lussac's law]1 Vol. CH4 requires 2 Vol. of O2∴ 10 cm3 CH4 will require 2 x 10= 20 cm3 of O2Given, air contains 20% O2 by volume.Let xxx volume of air contain 20 cm3 of O2⇒20100×x=20⇒x=10020×20⇒x=100 cm3\Rightarrow \dfrac{20}{100} \times x = 20 \[1em] \Rightarrow x = \dfrac{100}{20} \times 20 \[1em] \Rightarrow x = 100 \text{ cm}^3⇒10020​×x=20⇒x=20100​×20⇒x=100 cm3∴ 20 cm3 O2 is present in 100 cm3 of air.Similarly, 2 Vol C2H2 requires 5 Vol. of oxygen∴ 10 cm3 C2H2 will require 52\dfrac{5}{2}25​ x 10= 25 cm3 of oxygenGiven, air contains 20% O2 by volumeLet xxx volume of air contain 25 cm3 of O2⇒20100×x=25⇒x=10020×25⇒x=125 cm3\Rightarrow \dfrac{20}{100} \times x = 25 \[1em] \Rightarrow x = \dfrac{100}{20} \times 25 \[1em] \Rightarrow x = 125 \text{ cm}^3⇒10020​×x=25⇒x=20100​×25⇒x=125 cm3∴ 25 cm3 O2 is present in 125 cm3 of air.Hence, total volume of air required is 100 + 125 = 225 cm3

Question 3

Question 14LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to the atmosphere.C3H8 + 5O2 ⟶ 3CO2 + 4H2O2C4H10 + 13O2 ⟶ 8CO2 + 10H2O

Accepted Answer

Given, 10 litres of this mixture contains 60% propane and 40% butane. Hence, propane is 6 litres and butane is 4 litresC3H8+5O2⟶3CO2+4H2O1 vol.:5 vol.⟶3 vol.2C4H10+13O2⟶8CO2+10H2O2 vol.:13 vol.⟶8 vol.\begin{matrix} 	ext{C}_3	ext{H}_8 & + & 5	ext{O}_2 & \longrightarrow & 3	ext{CO}_2 & + & 4	ext{H}_2	ext{O} \ 1 	ext{ vol.} & : & 5 	ext{ vol.} & \longrightarrow & 3	ext{ vol.} \ 2	ext{C}_4	ext{H}_{10} & + & 13	ext{O}_2 & \longrightarrow & 8	ext{CO}_2 & + & 10	ext{H}_2	ext{O} \ 2 	ext{ vol.} & : & 13 	ext{ vol.} & \longrightarrow & 8	ext{ vol.} \ \end{matrix}C3​H8​1 vol.2C4​H10​2 vol.​+:+:​5O2​5 vol.13O2​13 vol.​⟶⟶⟶⟶​3CO2​3 vol.8CO2​8 vol.​++​4H2​O10H2​O​1 Vol. C3H8 produces carbon dioxide = 3 VolSo, 6 litres C3H8 will produce carbon dioxide = 3 x 6 = 18 litres2 Vol. C4H10 produces carbon dioxide = 8 VolSo, 4 litres C4H10 will produce carbon dioxide = 82\dfrac{8}{2}28​ x 4 = 16 litresHence, 34 (i.e., 18 + 16) litres of CO2 is produced.

Question 4

Question 15Water decomposes to O2 and H2 under suitable conditions as represented by the equation below:2H2O ⟶ 2H2 + O2(a) If 2500 cm3 of H2 is produced, what volume of O2 is liberated at the same time and under the same conditions of temperature and pressure?(b) The 2500 cm3 of H2 is subjected to 2122\dfrac{1}{2}221​ times increase in pressure (temp. remaining constant). What volume will H2 now occupy?(c) Taking the value of H2 calculated in 5(b), what changes must be made in Kelvin (absolute) temperature to return the volume to 2500 cm3 pressure remaining constant.

Accepted Answer

2H2O⟶2H2+O2 Vol.2 Vol.1 Vol.\begin{matrix} 2	ext{H}_2	ext{O} & \longrightarrow & 2	ext{H}_2 & + &	ext{O} \ 2	ext{ Vol.} & & 2	ext{ Vol.}& & 1	ext{ Vol.} \end{matrix}2H2​O2 Vol.​⟶​2H2​2 Vol.​+​O1 Vol.​2 Vol. of water gives 2 Vol. of H2 and 1 Vol. of O2∴ If 2500 cm3 of H2 is produced, volume of O2 produced = 25002\dfrac{2500}{2}22500​ = 1250 cm3(b) V1 = 2500 cm3P1 = 1 atm = 760 mmT1 = TT2 = TP2 = [760 x 2 12\dfrac{1}{2}21​] + [760] = 760 [52\dfrac{5}{2}25​ + 1] = 760 x 72\dfrac{7}{2}27​ = 2660 mmV2 = ?Using formula:P1V1T1\dfrac{	ext{P}_1	ext{V}_1}{	ext{T}_1}T1​P1​V1​​ = P2V2T2\dfrac{	ext{P}_2	ext{V}_2}{	ext{T}_2}T2​P2​V2​​760×2500T\dfrac{760 	imes 2500}{	ext{T}}T760×2500​ = 2660×V2T2660 	imes \dfrac{	ext{V}_2}{	ext{T}}2660×TV2​​V2 = 760×25002660\dfrac{760 	imes 2500 }{2660}2660760×2500​ = 50007\dfrac{5000}{7}75000​(c) V1 = 50007\dfrac{5000}{7}75000​ = 714.29 cm3P1 = P2 = PT1 = TV2 = 2500 cm3T2 = ?Using formula:P1V1T1\dfrac{	ext{P}_1	ext{V}_1}{	ext{T}_1}T1​P1​V1​​ = P2V2T2\dfrac{	ext{P}_2	ext{V}_2}{	ext{T}_2}T2​P2​V2​​P×714.29T\dfrac{	ext{P} 	imes 714.29}{	ext{T}}TP×714.29​ = P×2500T2\dfrac{	ext{P} 	imes 2500}{	ext{T}_2}T2​P×2500​T2 = 2500714.29\dfrac{2500 }{714.29}714.292500​ x TT2 = 3.5 x THence, T2 = 3.5 times T or temperature should be increased by 3.5 times

Question 5

Question 2Find the:(a) number of molecules in 73 g of HCl,(b) weight of 0.5 mole of O2,(c) number of molecules in 1.8 g of H2O,(d) number of moles in 10 g of CaCO3,(e) weight of 0.2 mole of H2 gas,(f) number of molecules in 3.2 g of SO2.

Accepted Answer

(a) Number of molecules in 73 g of HCl —Molecular wt. of any substance contain 6.022 × 1023 molecules.Mass of 1 mole of HCl is 1 + 35.5 = 36.5 g36.5 g of HCl contains 6.022 × 1023 molecules∴ 73 g of HCl contains 6.022×1023×7336.5\dfrac{6.022 	imes 10^{23} 	imes 73 }{36.5}36.56.022×1023×73​= 1.2 × 1024 molecules(b) Weight of 0.5 mole of O2 —1 mole of O2 weighs = 2O = 2 x 16 = 32 g∴ 0.5 moles will weigh = 322\dfrac{32}{2}232​ = 16 g(c) Number of molecules in 1.8 g of H2O —Molecular wt. of any substance contains 6.022 × 1023 molecules.Mass of 1 mole of H2O is (2 x 1) + 16 = 2 + 16 = 18 g18 g of H2O contains 6.022 × 1023 molecules∴ 1.8 g of H2O contains 6.022×1023×1.818\dfrac{6.022 	imes 10^{23} 	imes 1.8 }{18}186.022×1023×1.8​= 6.02 × 1022 molecules(d) Number of moles in 10 g of CaCO3 —Mass of 1 mole of CaCO3 = 40 + 12 + 3(16) = 52 + 48 = 100 g100 g of CaCO3 = 1 mole∴ 10 g of CaCO3 = 1×10100\dfrac{1 	imes 10}{100}1001×10​= 0.1 mole(e) Weight of 0.2 mole H2 gas —1 mole of H2 weighs = 2 g∴ 0.2 moles will weigh = 2×0.21\dfrac{2 	imes 0.2}{1}12×0.2​ = 0.4 g(f) No. of molecules in 3.2 g of SO2 —Molecular wt. of any substance contain 6 × 1023 molecules.Mass of 1 mole of SO2 is 32 + 2(16) = 32 + 32 = 64 g64 g of SO2 contains 6 × 1023 molecules∴ 3.2 g of SO2 contains 6×1023×3.264\dfrac{6 	imes 10^{23} 	imes 3.2 }{64}646×1023×3.2​= 3 x 1022 molecules.

Question 6

Question 15(a)Calculate the mass of nitrogen supplied to soil by 5 kg of urea [CO(NH2)2][O = 16; N = 14; C = 12 ; H = 1 ]

Accepted Answer

Molar mass of urea [CO(NH2)2] = 12 + 16 + 2(14 + (2 x 1))= 28 + 2(16)= 28 + 32= 60 gMolar mass of nitrogen = 2 x 14 = 28 g60 g urea has mass of nitrogen = 28 g∴ 5000 g urea will have mass= 28×500060\dfrac{28 	imes 5000 }{60}6028×5000​= 2333 g = 2.33 kg

Question 7

Question 15(b)Calculate the volume occupied by 320 g of sulphur dioxide at S.T.P.[S = 32; O = 16]

Accepted Answer

Molar mass of sulphur dioxide (SO2) = S + 2O = 32 + (2 x 16) = 32 + 32 = 64 g64 g of sulphur dioxide has volume = 22.4 litre∴ 320 g of sulphur dioxide will have volume = 22.4×32064\dfrac{22.4	imes 320}{64}6422.4×320​= 112 litres

Question 8

Question 16(a) What do you understand by the statement that 'vapour density of carbon dioxide is 22'?(b) Atomic mass of Chlorine is 35.5. What is it's vapour density?

Accepted Answer

(a) Vapour density of carbon dioxide is 22 implies that 1 molecule of carbon dioxide is 22 times heavier than 1 molecule of hydrogen.(b) Vapour density = Molecular mass2\dfrac{	ext{Molecular mass}}{2}2Molecular mass​Molecular mass of chlorine Cl2 = 2Cl = 2 x 35.5 = 71 gSubstituting in formula;Vapour density = 712\dfrac{71}{2}271​ = 35.5Hence, vapour density of Chlorine atom is 35.5

Question 9

Question 17What is the mass of 56 cm3 of carbon monoxide at S.T.P.?(C = 12, O = 16)

Accepted Answer

22400 cm3 of CO has mass = 12 + 16 = 28 g∴ 56 cm3 will have mass = 2822400\dfrac{28}{22400}2240028​ x 56 = 0.07 g

Question 10

Question 18Determine the number of molecules in a drop of water which weighs 0.09 g.

Accepted Answer

Molecular wt. of any substance contain 6.022 × 1023 molecules.Mass of 1 mole of water is 2H + O = 2 + 16 = 18 g18 g of H2O contains 6.022 × 1023 molecules∴ 0.09 g of H2O contains 6.022×1023×0.0918\dfrac{6.022 	imes 10^{23} 	imes 0.09 }{18}186.022×1023×0.09​= 3.01 × 1021 molecules

Question 11

Question 19The molecular formula for elemental sulphur is S8. In a sample of 5.12 g of sulphur:(a) How many moles of sulphur are present?(b) How many molecules and atoms are present?

Accepted Answer

(a) Mass of 1 mole of S8 = 8S = 8 x 32 = 256 g∴ Moles in 5.12 g of sulphur = 5.12256\dfrac{5.12}{256}2565.12​ = 0.02 moles(b) 1 mole = 6.022 × 1023 molecules∴ 0.02 moles will have = 0.02 × 6.022 × 1023= 1.2044 × 1022 ≈ 1.2 × 1022 moleculesNo. of atoms in 1 molecule of S8 = 8∴ No. of atoms in 1.2044 × 1022 molecules = 1.2044 x 1022 × 8= 9.635 × 1022 molecules

Question 12

Question 20If phosphorus is considered to contain P4 molecules, then calculate the number of moles in 100 g of phosphorus?

Accepted Answer

Mass of 1 mole of P4 = 4P = 4 x 31 = 124 g124 g of phosphorus (P4) = 1 mole∴ 100 g of phosphorus (P4) = 1124\dfrac{1}{124}1241​ x 100 = 0.806 moles

Question 13

Question 21Calculate:(a) The gram molecular mass of chlorine if 308 cm3 of it at S.T.P. weighs 0.979 g(b) The volume of 4 g of H2 at 4 atmospheres.(c) The mass of oxygen in 2.2 litres of CO2 at S.T.P.

Accepted Answer

(a) The mass of 22.4 L of a gas at S.T.P. is equal to it's gram molecular mass.308 cm3 of chlorine weighs 0.979 g∴ 22,400 cm3 of chlorine will weigh= 0.979308\dfrac{0.979}{308}3080.979​ × 22400 = 71.2 g(b) Molar mass of H2 = 2H = 2 x 1 = 2 g2g H2 at 1 atm has volume = 22.4 dm3∴ 4 g H2 at 1 atm will have volume 2 x 22.4 = 44.8 dm3Now, For 4 g H2P1 = 1 atm, V1 = 44.8 dm3P2 = 4 atm, V2 = ?Using formula P1V1 = P2V2V2=P1V1P2V2=1×44.84=11.2 dm3\text{V}_2 = \dfrac{\text{P}_1\text{V}_1}{\text{P}_2} \[1em] \text{V}_2 = \dfrac{1 \times 44.8}{4} \[1em] = \bold{11.2} \space \bold{dm^3}V2​=P2​P1​V1​​V2​=41×44.8​=11.2 dm3(c) Molar mass of oxygen in carbon dioxide = 2O = 2 x 16 = 32 gMass of oxygen in 22.4 litres of CO2 = 32 g∴ Mass of oxygen in 2.2 litres of CO2= 3222.4\dfrac{32}{22.4}22.432​ x 2.2 = 3.14 g

Question 14

Question 22A student puts his signature with graphite pencil. If the mass of carbon in the signature is 10-12 g, calculate the number of carbon atoms in the signature.

Accepted Answer

No. of atoms in 12 g C = 6.022 × 1023∴ no. of carbon atoms in 10-12 g6.022×102312\dfrac{6.022 	imes 10^{23}}{12}126.022×1023​ x 10-12= 5.019 × 1010 atoms

Question 15

Question 23An unknown gas shows a density of 3 g per litre at 273°C and 1140 mm Hg pressure. What is the gram molecular mass of this gas?

Accepted Answer

Given:P = 1140 mm HgDensity = D = 3 g per LT = 273 °C = 273 + 273 = 546 Kgram molecular mass = ?At S.T.P., the volume of one mole of any gas is 22.4 LVolume of unknown gas at S.T.P. = ?By Charle’s law.V1 = 1 LT1 = 546 KT2 = 273 KV2 = ?V1T1\dfrac{	ext{V}_1}{	ext{T}_1}T1​V1​​ = V2T2\dfrac{	ext{V}_2}{	ext{T}_2}T2​V2​​Hence, V2 = 1546\dfrac{1}{546}5461​ x 273 = 0.5 LVolume at standard pressure = ?Apply Boyle’s law.P1 = 1140 mm HgV1 = 0.5 LP2 = 760 mm HgV2 = ?P1 × V1 = P2 × V2V2 = 1140×0.5760\dfrac{1140 	imes 0.5}{760}7601140×0.5​ = 0.75 LNow,22.4 L = 1 mole of any gas at S.T.P.,then 0.75 L = 0.7522.4\dfrac{0.75}{22.4}22.40.75​= 0.0335 molesThe original mass is 3 gMolecular mass = Mass of compoundMoles of compound\dfrac{	ext{Mass of compound}}{	ext{Moles of compound}}Moles of compoundMass of compound​= 30.0335\dfrac{3}{0.0335 }0.03353​ = 89.55 ≈ 89.6 g per moleHence, the gram molecular mass of the unknown gas is 89.6g

Question 16

Question 24Cost of Sugar (C12H22O11) is ₹40 per kg; calculate it's cost per mole.

Accepted Answer

Molar mass of C12H22O11 = 12C + 22H + 11O = (12 x 12) + (22 x 1) + (11 x 16) = 144 + 22 + 176 = 342 g1000 g of sugar costs = ₹40∴ 342 g of sugar will cost = 401000\dfrac{40}{1000}100040​ x 342 = ₹13.68 per mole

Question 17

Question 25Calculate the number of molecules in one kg of NaOH.

Accepted Answer

Mass of 1 mole of NaOH = Na + O + H = 23 + 16 + 1 = 40 g40 g of NaOH contains 6.022 × 1023 molecules∴ 1000 g of NaOH contains = 6.022×1023×100040\dfrac{6.022 	imes 10^{23} 	imes 1000}{40}406.022×1023×1000​= 1.5 × 1025 molecules

Question 18

Question 26Calculate the number of atoms present in :(a) 10 g of Chlorine(b) 10 g of Nitrogen

Accepted Answer

(a) Mass of 1 mole of chlorine (Cl) is 35.5 g35.5 g of chlorine (Cl) contains 6.022 × 1023 atoms∴ 10g of chlorine (Cl) contains = 6.022×1023×1035.5\dfrac{6.022 	imes 10^{23} 	imes 10}{35.5}35.56.022×1023×10​= 1.7 x 1023 atoms(b) Mass of 1 mole of nitrogen (N) is 14 g14 g of nitrogen (N) contains 6.022 × 1023 atoms∴ 10g of nitrogen (N) contains = 6.022×1023×1014\dfrac{6.022 	imes 10^{23} 	imes 10}{14}146.022×1023×10​= 4.3 x 1023 atoms

Question 19

Question 27Correct the following:(a) Equal volumes of any gas, under similar conditions, contain an equal number of atoms.(b) 22 g of CO2, occupies 22.4 litres at STP.(c) The unit of atomic weight is grams.

Accepted Answer

(a) Equal volumes of any gas, under similar conditions, contain an equal number of molecules.(b) 44 g of CO2, occupies 22.4 litres at STP.(c) The unit of atomic weight is atomic mass unit (a.m.u).

Question 20

Question 11An organic compound contains 4.07% hydrogen, 71.65% chlorine and remaining carbon. Its molar mass is 98.96. Find its,(a) Empirical formula(b) Molecular formula

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioHydrogen4.0714.071\dfrac{4.07 }{1}14.07​ = 4.074.072.01\dfrac{4.07 }{2.01}2.014.07​ = 2chlorine71.6535.571.6535.5\dfrac{ 71.65}{35.5}35.571.65​ = 2.012.012.01\dfrac{2.01}{2.01 }2.012.01​ = 1Carbon24.281224.2812\dfrac{24.28 }{12}1224.28​ = 2.022.022.01\dfrac{ 2.02 }{2.01}2.012.02​ = 1Simplest ratio of whole numbers = H : Cl : C = 2 : 1 : 1Hence, empirical formula is CH2ClEmpirical formula weight = 12 + (2 x 1) + 35.5 = 49.5molar mass = 98.96n=Molecular weightEmpirical formula weight=98.9649.5=1.99=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{98.96}{49.5} = 1.99 = 2n=Empirical formula weightMolecular weight​=49.598.96​=1.99=2So, molecular formula = 2(CH2Cl) = C2H4Cl2

Mole Concept and Stoichiometry

Exercise 5A

Exercise 5B

Exercise 5C

Exercise 5D