Question 1

Question 1Nitrogen reacts with hydrogen to give ammonia. Calculate the volume of the ammonia gas formed when nitrogen reacts with 6 litres of hydrogen. All volumes measured at s.t.p.

Accepted Answer

[By Lussac's law]N2+3H2⟶2NH31 vol.:3 vol.⟶2 vol.\begin{matrix} \text{N}_2 & + & 3\text{H}_2 & \longrightarrow & 2 \text{NH}_3 \ 1 \text{ vol.} & : & 3 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}N2​1 vol.​+:​3H2​3 vol.​⟶⟶​2NH3​2 vol.​To calculate the volume of ammonia gas formed.H2:NH33:26:x\begin{matrix} \text{H}_2 & : & \text{NH}_3 \ 3 & : & 2 \ 6 & : & x \end{matrix}H2​36​:::​NH3​2x​Therefore,23×6=x⇒x=4 lts\dfrac{2}{3} \times 6 = x \[0.5em] \Rightarrow x = 4 \text{ lts}32​×6=x⇒x=4 ltsHence, volume of ammonia gas formed is 4 lts

Question 2

Question 22500 cc of oxygen was burnt with 600 cc of ethane [C2H6]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed.

Accepted Answer

[By Lussac's law]2C2H6+7O2⟶4CO2+6H2O2 vol.:7 vol.⟶4 vol.:6 vol.\begin{matrix} \text{\scriptsize{2C}}_2\text{\scriptsize{H}}_6 & \text{\scriptsize{+}} & \text{\scriptsize{7O}}_2 & \longrightarrow & \text{\scriptsize{4CO}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{6H}}_2\text{\scriptsize{O}} \ \text{\scriptsize{2 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{7 vol.}} & \longrightarrow & \text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{6 vol.}} \end{matrix}2C2​H6​2 vol.​+:​7O2​7 vol.​⟶⟶​4CO2​4 vol.​+:​6H2​O6 vol.​To calculate the volume of unused oxygen.C2H6:O22:7600:x\begin{matrix} \text{C}_2\text{H}_6 & : &\text{O}_2 \ 2 & : & 7 \ 600 & : & x \end{matrix}C2​H6​2600​:::​O2​7x​Therefore,72×600=x⇒x=2100cc\dfrac{7}{2} \times 600 = x \[0.5em] \Rightarrow x = 2100 \text{cc}27​×600=x⇒x=2100ccTherefore, volume of unused oxygen = 2500 - 2100 = 400 ccTo calculate the volume of carbon dioxide formedC2H6:CO22:4600:x\begin{matrix} \text{C}_2\text{H}_6 & : & \text{CO}_2 \ 2 & : & 4 \ 600 & : & x \end{matrix}C2​H6​2600​:::​CO2​4x​Therefore,42×600=x⇒x=1200cc\dfrac{4}{2} \times 600 = x \[0.5em] \Rightarrow x = 1200 \text{cc}24​×600=x⇒x=1200ccTherefore, volume of carbon dioxide formed is 1200 cc

Question 3

Question 320 ml. each of oxygen and hydrogen and 10 ml. of carbon monoxide are exploded in an enclosure. What will be the volume and composition of the mixture of the gases when they are cooled to room temperature.

Accepted Answer

Given,20 ml. each of oxygen and hydrogen and 10 ml. of carbon monoxide[By Lussac's law]2CO+O2⟶2CO22 vol.:1 vol.⟶2 vol.\begin{matrix} 2\text{CO}& + & \text{O}_2 & \longrightarrow & 2\text{CO}_2 & \ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} & \ \end{matrix}2CO2 vol.​+:​O2​1 vol.​⟶⟶​2CO2​2 vol.​​To calculate the amount of CO2 produced,CO:CO22 vol.:2 vol.10 ml:x ml\begin{matrix} \text{CO}& : & \text{CO}_2 & \ 2 \text{ vol.} & : & 2 \text{ vol.} & \ 10 \text{ ml} & : & x \text{ ml} & \ \end{matrix}CO2 vol.10 ml​:::​CO2​2 vol.x ml​​Therefore, CO2 produced is 10 mlTo calculate the amount of O2 used,CO:O22 vol.:1 vol.10 ml:x ml\begin{matrix} \text{CO}& : & \text{O}_2 & \ 2 \text{ vol.} & : & 1 \text{ vol.} & \ 10 \text{ ml} & : & x \text{ ml} & \ \end{matrix}CO2 vol.10 ml​:::​O2​1 vol.x ml​​12×10=x⇒x=5 ml\dfrac{1}{2} \times 10 = x \[0.5em] \Rightarrow x = 5 \text{ ml}21​×10=x⇒x=5 mlFrom relation,2H2+O2⟶2H2O2 vol.:1 vol.⟶2 vol.\begin{matrix} 2\text{H}_2 & + & \text{O}_2 & \longrightarrow & 2\text{H}_2 \text {O} & \ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} & \ \end{matrix}2H2​2 vol.​+:​O2​1 vol.​⟶⟶​2H2​O2 vol.​​H2:O22 vol.:1 vol.20 ml:x ml\begin{matrix} \text{H}_2 & : & \text{O}_2 & \ 2 \text{ vol.} & : & 1 \text{ vol.} & \ 20 \text{ ml} & : & x \text{ ml} & \ \end{matrix}H2​2 vol.20 ml​:::​O2​1 vol.x ml​​12×20=x⇒x=10 ml\dfrac{1}{2} \times 20 = x \[0.5em] \Rightarrow x = 10 \text{ ml}21​×20=x⇒x=10 mlTherefore, total vol. of oxygen used = 10 + 5 = 15 mlHence, oxygen left = 20 - 15 = 5 mlTherefore, oxygen left is 5 ml and CO2 produced is 10 ml.

Question 4

Question 4224 cm3 of ammonia undergoes catalytic oxidation in presence of Pt to given nitric oxide and water vapour. Calculate the volume of oxygen required for the reaction. All volumes measured at room temperature and pressure.

Accepted Answer

[By Lussac's law]4NH3+5O2⟶4NO+6H2O4 vol.:5 vol.⟶4 vol.:6 vol.\begin{matrix} \text{\scriptsize{4NH}}_3 & \text{\scriptsize{+}} & \text{\scriptsize{5O}}_2 & \longrightarrow & \text{\scriptsize{4NO}} & \text{\scriptsize{+}} & \text{\scriptsize{6H}}_2\text{\scriptsize{O}} \ \text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{5 vol.}} & \longrightarrow &\text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{6 vol.}} \end{matrix}4NH3​4 vol.​+:​5O2​5 vol.​⟶⟶​4NO4 vol.​+:​6H2​O6 vol.​To calculate the volume of oxygen required.NH3:O24:5224:x\begin{matrix} \text{NH}_3 & : & \text{O}_2 & \ 4 & : & 5 \ 224 & : & x \end{matrix}NH3​4224​:::​O2​5x​Therefore,54×224=x⇒x=280 cm3\dfrac{5}{4} \times 224 = x \[0.5em] \Rightarrow x = 280 \text{ cm}^345​×224=x⇒x=280 cm3Therefore, volume of oxygen required is 280 cm3.

Question 5

Question 5Acetylene [C2H2] burns in air forming carbon dioxide and water vapour. Calculate the volume of air required to completely burn 50 cm3 of acetylene. [Assume air contains 20% oxygen].

Accepted Answer

[By Lussac's law]2C2H2+5O2⟶4CO2+2H2O2 vol.:5 vol.⟶4 vol.:2 vol.\begin{matrix} \text{\scriptsize{2C}}_2 \text{\scriptsize{H}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{5O}}_2 & \longrightarrow & \text{\scriptsize{4CO}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{2H}}_2\text{\scriptsize{O}} \ \text{\scriptsize{2 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{5 vol.}} & \longrightarrow & \text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{2 vol.}} \end{matrix}2C2​H2​2 vol.​+:​5O2​5 vol.​⟶⟶​4CO2​4 vol.​+:​2H2​O2 vol.​To calculate the volume of air required,C2H2:O22:550:x\begin{matrix} \text{C}_2 \text{H}_2 & : & \text{O}_2 & \ 2 & : & 5 \ 50 & : & x \end{matrix}C2​H2​250​:::​O2​5x​Therefore, volume of oxygen (x),52×50=x⇒x=125 cm3\dfrac{5}{2} \times 50 = x \[0.5em] \Rightarrow x = 125 \text{ cm}^325​×50=x⇒x=125 cm3When oxygen is 20% then air is 100%Therefore when, oxygen is 125 cm3 then air is10020×125=x⇒x=625 cm3\dfrac{100}{20} \times 125 = x \[0.5em] \Rightarrow x = 625 \text{ cm}^320100​×125=x⇒x=625 cm3Therefore, volume of air required is 625 cm3.

Question 6

Question 6On igniting a mixture of acetylene [C2H2] and oxygen, 200 cm3 of CO2 is collected at s.t.p. Calculate the volume of acetylene & O2 at s.t.p. in the original mixture.

Accepted Answer

[By Lussac's law]2C2H2+5O2⟶4CO2+2H2O2 vol.:5 vol.⟶4 vol.:2 vol.\begin{matrix} \text{\scriptsize{2C}}_2 \text{\scriptsize{H}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{5O}}_2 & \longrightarrow & \text{\scriptsize{4CO}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{2H}}_2\text{\scriptsize{O}} \ \text{\scriptsize{2 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{5 vol.}} & \longrightarrow & \text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{2 vol.}} \end{matrix}2C2​H2​2 vol.​+:​5O2​5 vol.​⟶⟶​4CO2​4 vol.​+:​2H2​O2 vol.​To calculate the volume of acetylene :CO2:C2H24:2200:x\begin{matrix} \text{CO}_2 & : & \text{C}_2 \text{H}_2 & \ 4 & : & 2 \ 200 & : & x \end{matrix}CO2​4200​:::​C2​H2​2x​Therefore,24×200=x⇒x=100 cm3\dfrac{2}{4} \times 200 = x \[0.5em] \Rightarrow x = 100 \text { cm}^342​×200=x⇒x=100 cm3Hence, volume of acetylene in the original mixture is 100 cm3.To calculate the volume of oxygen :CO2:O24:5200:x\begin{matrix} \text{CO}_2 & : & \text{O}_2 & \ 4 & : & 5 \ 200 & : & x \end{matrix}CO2​4200​:::​O2​5x​Therefore,54×200=x⇒x=250 cm3\dfrac{5}{4} \times 200 = x \[0.5em] \Rightarrow x = 250 \text { cm}^345​×200=x⇒x=250 cm3Hence, volume of oxygen in the original mixture is 250 cm3.

Question 7

Question 7Ammonia is formed from the reactants nitrogen and hydrogen in presence of a catalyst under suitable conditions. Assuming all volumes are measured in litres at s.t.p. Calculate the volume of ammonia formed if only 10% conversion has taken place.

Accepted Answer

[By Lussac's law]N2+3H2⟶2NH31 vol.:3 vol.⟶2 vol.1 lit.:3 lit.⟶2 lit.\begin{matrix} 	ext{N}_2 & + & 3	ext{H}_2 & \longrightarrow & 2	ext{NH}_3 & \ 1 	ext{ vol.} & : & 3 	ext{ vol.} & \longrightarrow & 2 	ext{ vol.} & \ 1 	ext{ lit.} & : & 3 	ext{ lit.} & \longrightarrow & 2 	ext{ lit.} & \end{matrix}N2​1 vol.1 lit.​+::​3H2​3 vol.3 lit.​⟶⟶⟶​2NH3​2 vol.2 lit.​​To calculate the volume of ammonia formed when only 10% conversion takes place,∴10% of 2 lit=10100×2=0.2 lit	herefore 10\% 	ext{ of 2 lit} = \dfrac{10}{100} 	imes 2 = 0.2 	ext{ lit}∴10% of 2 lit=10010​×2=0.2 litHence, volume of ammonia formed is 0.2 lit or 20% or 1/5th of vol of N2 and H2

Question 8

Question 8100 cc. each of water gas and oxygen are ignited and the resultant mixture of gases cooled to room temp. Calculate the composition of the resultant mixture. [Water gas contains CO and H2 in equal ratio]

Accepted Answer

Given,Vol of CO = 50 cc and H2 = 50 cc [As Water gas contains CO and H2 in equal ratio] and O2 = 100 cc[By Lussac's law]CO1 vol.+:H21 vol.+:O21 vol.⟶CO21 vol +: H2O1 vol\underset{	ext{\scriptsize{1 vol.}}}{	ext{\scriptsize{CO}}} \underset{:}{	ext{\scriptsize{+}}} \underset{	ext{\scriptsize{1 vol.}}}{	ext{\scriptsize{H}}_2} \underset{:}{	ext{\scriptsize{+}}} \underset{	ext{\scriptsize{1 vol.}}}{	ext{\scriptsize{O}}_2} \longrightarrow \underset{	ext{\scriptsize{1 vol}}}{	ext{\scriptsize{CO}}_2} \underset{:}{	ext{\scriptsize{ +}}} \underset{	ext{\scriptsize{1 vol}}}{	ext{\scriptsize{ H}}_2 	ext{\scriptsize{O}}}1 vol.CO​:+​1 vol.H2​​:+​1 vol.O2​​⟶1 volCO2​​: +​1 vol H2​O​To calculate the amount of O2 used,CO:O21 vol.:1 vol.50 cc:x cc\begin{matrix} 	ext{CO}& : & 	ext{O}_2 & \ 1 	ext{ vol.} & : & 1 	ext{ vol.} & \ 50 	ext{ cc} & : & x 	ext{ cc} & \ \end{matrix}CO1 vol.50 cc​:::​O2​1 vol.x cc​​As ratio between CO and O2 is same so, O2 used is 50 cc.Hence, remaining O2 = 100 - 50 = 50 cc.To calculate the amount of CO2 produced,CO:CO21 vol.:1 vol.50 cc:y cc\begin{matrix} 	ext{CO}& : & 	ext{CO}_2 & \ 1 	ext{ vol.} & : & 1 	ext{ vol.} & \ 50 	ext{ cc} & : & y 	ext{ cc} & \ \end{matrix}CO1 vol.50 cc​:::​CO2​1 vol.y cc​​1 vol of CO produces 1 vol. of CO2Hence, CO2 produced = 50 cc.Therefore, the resultant mixture has 50 cc of O2 + 50 cc of CO2

Question 9

Question 9Calculate the number of molecules in 1½ litres of water. [density of water 1.0 g./cc. ∴ mass of water = volume × density]

Accepted Answer

Given,Vol of water = 1.5 lit = 1500 mL =1500 ccDensity of water 1.0 g./cc..∴ Mass of water = volume × densityHence, mass of water = 1500 x 1 = 1500 gGram molecular mass of water= 2H + O= (2 x 1) + 16 = 18 gSo, 18 g of water = 6.023 x 1023 moleculesTherefore, 1500 g of water will have6.023×102318×1500=150018×6.023×1023=83.33×6.023×1023\dfrac{6.023 \times 10^{23}}{18} \times 1500 \[0.5em] = \dfrac{1500}{18} \times 6.023 \times 10^{23} \[0.5em] = 83.33 \times 6.023 \times 10^{23}186.023×1023​×1500=181500​×6.023×1023=83.33×6.023×1023Hence, the number of molecules in 1½ litres of water is 83.33 x 6.023 x 1023 molecules.

Question 10

Question 10Calculate the gram-atoms in 88.75 g of chlorine [Cl = 35.5]

Accepted Answer

Gram atoms is the relative atomic mass of an element expressed in grams.Gram atoms=Mass in gramsRelative At. mass [At. wt.]=88.7535.5=2.5 gram atoms\text{Gram atoms} = \dfrac{\text{Mass in grams}}{\text{Relative At. mass [At. wt.]}} \[0.5em] = \dfrac{88.75}{35.5} \[0.5em] = 2.5 \text { gram atoms} \[0.5em]Gram atoms=Relative At. mass [At. wt.]Mass in grams​=35.588.75​=2.5 gram atomsHence, gram-atoms in 88.75 g of chlorine is 2.5 g. atoms

Question 11

Question 11Calculate the number of hydrogen atoms in 0.25 mole of H2SO4.

Accepted Answer

1 mole of hydrogen atom has 2 x 6.023 x 1023 atoms.∴ 0.25 mole will have=2×6.023×10231×0.25=0.5×6.023×1023 atoms= \dfrac{2 \times 6.023 \times 10^{23}}{1} \times 0.25 \[0.5em] = 0.5 \times 6.023 \times 10^{23} \text{ atoms}=12×6.023×1023​×0.25=0.5×6.023×1023 atomsHence, number of hydrogen atoms in 0.25 mole of H2SO4 = 0.5 x 6.023 x 1023 particles

Question 12

Question 12Calculate the gram molecules in 21 g of nitrogen [N = 14]

Accepted Answer

Gram molecules is the relative molecular mass of a substance expressed in grams.Relative molecular mass of N2 = 2 x 14 = 28 gGram molecules=Mass in grams [of nitrogen]Rel. molecular mass [Mol. wt.]=2128=0.75 gram molecules\text{Gram molecules} = \dfrac{\text{Mass in grams [of nitrogen]}}{\text{Rel. molecular mass [Mol. wt.]}} \[0.5em] = \dfrac{21}{28} \[0.5em] = 0.75 \text { gram molecules}Gram molecules=Rel. molecular mass [Mol. wt.]Mass in grams [of nitrogen]​=2821​=0.75 gram moleculesHence, gram molecules in 21 g of nitrogen = 0.75 gram molecules

Question 13

Question 13Calculate the number of atoms in 10 litres of ammonia [N = 14, H = 1]

Accepted Answer

1 mole of NH3 = N + 3 H = 4 atoms = 4 x 6.023 x 1023 atoms and occupies 22.4 lit at s.t.p.If 22.4 lit of NH3 has 4 x 6.023 x 1023 atoms,then, 10 lit will have=4×6.023×102322.4×10=4022.4×6.023×1023=1.786×6.023×1023 atoms= \dfrac{4 \times 6.023 \times 10^{23}}{22.4} \times 10 \[0.5em] = \dfrac{40}{22.4} \times 6.023 \times 10^{23} \[0.5em] = 1.786 \times 6.023 \times 10^{23} \text{ atoms}=22.44×6.023×1023​×10=22.440​×6.023×1023=1.786×6.023×1023 atomsHence, number of atoms in 10 litres of ammonia = 1.786 x 6.023 x 1023 atoms

Question 14

Question 14Calculate the number of atoms in 60 g of neon [Ne = 20]

Accepted Answer

Gram atomic mass of Ne = 20 gAs, 1 mole of Ne weighs 20 g and has 6.023 x 1023 atomsSo, 60 g of Ne will have6.023×102320×60=6020×6.023×1023=3×6.023×1023 atoms\dfrac{ 6.023 \times 10^{23}}{20} \times 60 \[0.5em] = \dfrac{60}{20} \times 6.023 \times 10^{23} \[0.5em] = 3 \times 6.023 \times 10^{23} \text{ atoms}206.023×1023​×60=2060​×6.023×1023=3×6.023×1023 atomsHence, number of atoms in 60 g of Ne = 3 x 6.023 x 1023 atoms

Question 15

Question 15Calculate the number of moles of 'X' atoms in 93 g of 'X' [X is phosphorus = 31]

Accepted Answer

Gram atomic mass of X (phosphorus) = 31 gAs, 31 g of phosphorus = 1 moleSo, 93 g of phosphorus = 131×93=3 moles\dfrac{1}{31} 	imes 93 = 3 	ext{ moles}311​×93=3 molesHence, number of moles in 93 g of X is 3 moles

Question 16

Question 16Calculate the volume occupied by 3.5 g of O2 gas at 27 °C and 740 mm pressure. [O = 16]

Accepted Answer

Gram molecular mass of O2 = 2 x 16 = 32 g1 mole of O2 weighs 32 g and occupies 22.4 lit. vol.∴ 3.5 g of O2 occupies = 22.432×3.5=2.45 lit.\dfrac{22.4}{32} \times 3.5 = 2.45 \text{ lit.}3222.4​×3.5=2.45 lit.Volume occupied by 3.5 g of O2 gas at 27°C and 740 mm pressure:s.t.p.given valuesP1 = 760 mm of HgP2 = 740 mm of HgV1 = 2.45 litV2 = x litT1 = 273 KT2 = 27 + 273 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,760×2.45273=740×x300x=760×2.45×300740×273x=5,58,6002,02,020x=2.76 lit\dfrac{760 \times 2.45}{273} = \dfrac{740 \times x}{300} \[0.5em] x = \dfrac{760 \times 2.45 \times 300}{740 \times 273 } \[0.5em] x = \dfrac{5,58,600}{2,02,020} \ \[0.5em] x = 2.76 \text{ lit}273760×2.45​=300740×x​x=740×273760×2.45×300​x=2,02,0205,58,600​x=2.76 litHence, the volume occupied by 3.5 g of O2 gas at 27°C and 740 mm pressure is 2.76 lit.

Question 17

Question 17Calculate the moles of sodium hydroxide contained in 160 g of it. [Na = 23, O = 16, H = 1]

Accepted Answer

Gram molecular mass of sodium hydroxide= Na + O + H= 23 + 16 + 1= 40 gAs, 40 g of sodium hydroxide = 1 mole∴ 160 g of sodium hydroxide = 140×160=4 moles\dfrac{ 1}{40} 	imes 160 = 4 	ext{ moles}401​×160=4 molesHence, number of moles in 160 g of sodium hydroxide is 4

Question 18

Question 18Calculate the weight in g. of 2.5 moles of ethane [C2H6]. [C = 12, H = 1]

Accepted Answer

Gram molecular mass of C2H6 = (2 x 12) + (6 x 1)= 24 + 6= 30 g1 mole of C2H6 weighs 30 g∴ 2.5 moles weighs = 301×2.5=75 g\dfrac{ 30}{1} 	imes 2.5 = 75 	ext{ g}130​×2.5=75 gHence, weight of 2.5 moles of ethane [C2H6] is 75 g.

Question 19

Question 19Calculate the molecular weight of 2.6 g of a gas which occupies 2.24 lits. at 0°C and 760 mm press.

Accepted Answer

Given,2.24 lit volume of gas weighs 2.6 g∴ 22.4 lit volume of gas weighs = 2.62.24×22.4=26 g\dfrac{2.6}{2.24} 	imes 22.4 = 26 	ext{ g}2.242.6​×22.4=26 gAs 22.4 lit of gas at s.t.p. weighs 26 g∴ Molecular weight = 26 g

Question 20

Question 20Calculate the gram atoms in 46 g of sodium [Na = 23]

Accepted Answer

Gram atoms is the relative atomic mass of an element expressed in grams.Gram atoms=Mass in grams [of sodium]Rel. atomic mass [At. wt.]=4623=2 gram atoms\text{Gram atoms} = \dfrac{\text{Mass in grams [of sodium]}}{\text{Rel. atomic mass [At. wt.]}} \[0.5em] = \dfrac{46}{23} = 2 \text { gram atoms}Gram atoms=Rel. atomic mass [At. wt.]Mass in grams [of sodium]​=2346​=2 gram atomsHence, gram-atoms in 46 g of sodium is 2 g. atoms

Gay Lussac's Law — Avogadro's Law — Mole Concept

Lussac's Law

Mole Concept — Avogadro's Law and No.