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Chapter 4A

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

Class 10 - Dalal Simplified ICSE Chemistry Solutions


Lussac's Law

Question 1

Nitrogen 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.

Answer

[By Lussac's law]

N2+3H22NH31 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}

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}

Therefore,

23×6=xx=4 lts\dfrac{2}{3} \times 6 = x \\[0.5em] \Rightarrow x = 4 \text{ lts}

Hence, volume of ammonia gas formed is 4 lts

Question 2

2500 cc of oxygen was burnt with 600 cc of ethane [C2H6]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed.

Answer

[By Lussac's law]

2C2H6+7O24CO2+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}

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}

Therefore,

72×600=xx=2100cc\dfrac{7}{2} \times 600 = x \\[0.5em] \Rightarrow x = 2100 \text{cc}

Therefore, volume of unused oxygen = 2500 - 2100 = 400 cc

To calculate the volume of carbon dioxide formed

C2H6:CO22:4600:x\begin{matrix} \text{C}_2\text{H}_6 & : & \text{CO}_2 \\ 2 & : & 4 \\ 600 & : & x \end{matrix}

Therefore,

42×600=xx=1200cc\dfrac{4}{2} \times 600 = x \\[0.5em] \Rightarrow x = 1200 \text{cc}

Therefore, volume of carbon dioxide formed is 1200 cc

Question 3

20 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.

Answer

Given,

20 ml. each of oxygen and hydrogen and 10 ml. of carbon monoxide

[By Lussac's law]

2CO+O22CO22 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}

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}

Therefore, CO2 produced is 10 ml

To 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}

12×10=xx=5 ml\dfrac{1}{2} \times 10 = x \\[0.5em] \Rightarrow x = 5 \text{ ml}

From relation,

2H2+O22H2O2 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}

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}

12×20=xx=10 ml\dfrac{1}{2} \times 20 = x \\[0.5em] \Rightarrow x = 10 \text{ ml}

Therefore, total vol. of oxygen used = 10 + 5 = 15 ml

Hence, oxygen left = 20 - 15 = 5 ml

Therefore, oxygen left is 5 ml and CO2 produced is 10 ml.

Question 4

224 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.

Answer

[By Lussac's law]

4NH3+5O24NO+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}

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}

Therefore,

54×224=xx=280 cm3\dfrac{5}{4} \times 224 = x \\[0.5em] \Rightarrow x = 280 \text{ cm}^3

Therefore, volume of oxygen required is 280 cm3.

Question 5

Acetylene [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].

Answer

[By Lussac's law]

2C2H2+5O24CO2+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}

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}

Therefore, volume of oxygen (x),

52×50=xx=125 cm3\dfrac{5}{2} \times 50 = x \\[0.5em] \Rightarrow x = 125 \text{ cm}^3

When oxygen is 20% then air is 100%
Therefore when, oxygen is 125 cm3 then air is

10020×125=xx=625 cm3\dfrac{100}{20} \times 125 = x \\[0.5em] \Rightarrow x = 625 \text{ cm}^3

Therefore, volume of air required is 625 cm3.

Question 6

On 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.

Answer

[By Lussac's law]

2C2H2+5O24CO2+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}

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}

Therefore,

24×200=xx=100 cm3\dfrac{2}{4} \times 200 = x \\[0.5em] \Rightarrow x = 100 \text { cm}^3

Hence, 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}

Therefore,

54×200=xx=250 cm3\dfrac{5}{4} \times 200 = x \\[0.5em] \Rightarrow x = 250 \text { cm}^3

Hence, volume of oxygen in the original mixture is 250 cm3.

Question 7

Ammonia 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.

Answer

[By Lussac's law]

N2+3H22NH31 vol.:3 vol.2 vol.1 lit.:3 lit.2 lit.\begin{matrix} \text{N}_2 & + & 3\text{H}_2 & \longrightarrow & 2\text{NH}_3 & \\ 1 \text{ vol.} & : & 3 \text{ vol.} & \longrightarrow & 2 \text{ vol.} & \\ 1 \text{ lit.} & : & 3 \text{ lit.} & \longrightarrow & 2 \text{ lit.} & \end{matrix}

To calculate the volume of ammonia formed when only 10% conversion takes place,

10% of 2 lit=10100×2=0.2 lit\therefore 10\% \text{ of 2 lit} = \dfrac{10}{100} \times 2 = 0.2 \text{ lit}

Hence, volume of ammonia formed is 0.2 lit or 20% or 1/5th of vol of N2 and H2

Question 8

100 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]

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{\text{\scriptsize{1 vol.}}}{\text{\scriptsize{CO}}} \underset{:}{\text{\scriptsize{+}}} \underset{\text{\scriptsize{1 vol.}}}{\text{\scriptsize{H}}_2} \underset{:}{\text{\scriptsize{+}}} \underset{\text{\scriptsize{1 vol.}}}{\text{\scriptsize{O}}_2} \longrightarrow \underset{\text{\scriptsize{1 vol}}}{\text{\scriptsize{CO}}_2} \underset{:}{\text{\scriptsize{ +}}} \underset{\text{\scriptsize{1 vol}}}{\text{\scriptsize{ H}}_2 \text{\scriptsize{O}}}

To calculate the amount of O2 used,

CO:O21 vol.:1 vol.50 cc:x cc\begin{matrix} \text{CO}& : & \text{O}_2 & \\ 1 \text{ vol.} & : & 1 \text{ vol.} & \\ 50 \text{ cc} & : & x \text{ cc} & \\ \end{matrix}

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} \text{CO}& : & \text{CO}_2 & \\ 1 \text{ vol.} & : & 1 \text{ vol.} & \\ 50 \text{ cc} & : & y \text{ cc} & \\ \end{matrix}

1 vol of CO produces 1 vol. of CO2
Hence, CO2 produced = 50 cc.

Therefore, the resultant mixture has 50 cc of O2 + 50 cc of CO2

Mole Concept — Avogadro's Law and No.

Question 1

Calculate the mass of 2.8 litres of CO2. [C = 12, O = 16]

Answer

[1 mole = 1 gm mol. wt. and occupies 22.4 lit at s.t.p.]

gm mol. wt. of CO2 = 12 + (16 x 2) = 44 g

1 mole of CO2 = 1 gm mol. wt.

22.4 lit of CO2 has mass 44 g [s.t.p.]

Therefore, 2.8 lit of CO2 has mass

=4422.4×2.8=5.5 g= \dfrac{44}{22.4} \times 2.8 = 5.5 \text{ g}

Therefore, mass of 2.8 litres of CO2 is 5.5 g

Question 2

Calculate the volume occupied by 53.5 g of Cl2. [Cl = 35.5]

Answer

[1 mole = 1 gm mol. wt. and occupies 22.4 lit at s.t.p.]

gm mol. wt. of Cl2 = (35.5 x 2) = 71 g

1 mole of Cl2 = 1 gm mol. wt. and occupies 22.4 lit.

71 g of Cl2 occupies 22.4 lit [s.t.p.]

∴ Vol. occupied by 53.5 g of Cl2

=22.471×53.5=16.87 lit= \dfrac{22.4}{71} \times 53.5 = 16.87 \text{ lit}

Therefore, volume occupied by 53.5 g of Cl2 is 16.87 lit

Question 3

Calculate the number of molecules in 109.5 g of HCl. [H = 1, Cl = 35.5]

Answer

gm mol. wt. of HCl = 1 + 35.5 = 36.5 g

At s.t.p.,

36.5 g of HCl = 6.023 x 1023 number of molecules    [Avogadro's law]

Therefore, 109.5 g of HCl

=6.023×102336.5×109.5=109.536.5×6.023×1023=3×6.023×1023= \dfrac{6.023 \times 10^{23}}{36.5} \times 109.5 \\[0.5em] = \dfrac{109.5}{36.5} \times 6.023 \times 10^{23} \\[0.5em] = 3 \times 6.023 \times 10^{23} \\[0.5em]

Hence, number of molecules in 109.5 g of HCl is 3 x 6.023 x 1023

Question 4

Calculate the number of :

(i) molecules [S = 32]

(ii) atoms in 192 g. of sulphur. [S8]

Answer

(i) 1 mole of any substance contains 6.023 x 1023 number of molecules.

S8 = 8 atoms = 8 x 32 = 256.

256 g = 6.023 x 1023 number of molecules

So, 192 g will have

=6.023×1023256×192=192256×6.023×1023=0.75×6.023×1023 molecules= \dfrac{6.023 \times 10^{23}}{256} \times 192 \\[0.5em] = \dfrac{192}{256} \times 6.023 \times 10^{23} \\[0.5em] = 0.75 \times 6.023 \times 10^{23} \text { molecules} \\[0.5em]

(ii) Gram atomic mass of S = 32 g

32 g of S has 6.023 x 1023 number of atoms

Therefore, 192 g will have

=6.023×102332×192=19232×6.023×1023=6×6.023×1023 atoms= \dfrac{6.023 \times 10^{23}}{32} \times 192 \\[0.5em] = \dfrac{192}{32} \times 6.023 \times 10^{23} \\[0.5em] = 6 \times 6.023 \times 10^{23} \text { atoms} \\[0.5em]

Question 5

Calculate the mass of Na which will contain 6.023 × 1023 atoms. [Na = 23]

Answer

As, a mole of atoms contain 6.023 × 1023 atoms [Avogadro number] and has weight [mass] equal to gram atomic mass of the element.

Therefore, a mole of Na will contain 6.023 × 1023 atoms and will have mass equal to gram atomic mass of Na = 23 g

Question 6

Calculate the no. of atoms of potassium present in 117 g. of K. [K = 39]

Answer

Gram atomic mass of K = 39

At s.t.p.,

39 g of K = 6.023 x 1023 atoms of K    [Avogadro's law]

Therefore, 117 g of K

=6.023×102339×117=11739×6.023×1023=3×6.023×1023= \dfrac{6.023 \times 10^{23}}{39} \times 117 \\[0.5em] = \dfrac{117}{39} \times 6.023 \times 10^{23} \\[0.5em] = 3 \times 6.023 \times 10^{23} \\[0.5em]

Hence, number of atoms in 117 g of K = 3 x 6.023 x 1023

Question 7

Calculate the number of moles and molecules in 19.86 g. of Pb(NO3)2. [Pb = 207, N = 14, O = 16]

Answer

Gram molecular mass of Pb(NO3)2
= Pb + 2 [N x 3(O)]
= Pb + 2N + 6O
= 207 + (2 x 14) + (6 x 16)
= 207 + 28 + 96 = 331

As,

331 g of Pb(NO3)2 = 1 mole
Therefore, 19.86 g

=1331×19.86=0.06 moles= \dfrac{1}{331} \times 19.86 = 0.06 \text { moles} \\[0.5em]

Hence, number of moles in 19.86 g. of Pb(NO3)2 = 0.06 moles

As,

1 mole of Pb(NO3)2 weighs 331 g and has 6.023 x 1023 molecules

Therefore, 19.86 g of Pb(NO3)2 will have

=6.023×1023331×19.86=19.86331×6.023×1023=0.06×6.023×1023molecules= \dfrac{ 6.023 \times 10^{23}}{331} \times 19.86 \\[0.5em] = \dfrac{19.86}{331} \times 6.023 \times 10^{23} \\[0.5em] = 0.06 \times 6.023 \times 10^{23} \text{molecules}

Hence, number of molecules in 19.86 g. of Pb(NO3)2 = 0.06 x 6.023 x 1023 moles

Question 8

Calculate the mass of an atom of lead [Pb = 202].

Answer

As,

1 mole of Pb weighs 202 g and has 6.023 x 1023 atoms.

So, 6.023 x 1023 atoms of Pb has mass = 202 g

Therefore, 1 atom of Pb will have mass

=2026.023×1023=33.53×1023g= \dfrac{202}{6.023 \times 10^{23}} = 33.53 \times 10^{-23} \text{g}

Hence, mass of an atom of lead is 33.53 x 10-23 g

Question 9

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

Answer

Given,

Vol of water = 1.5 lit = 1500 mL =1500 cc

Density of water 1.0 g./cc..

∴ Mass of water = volume × density

Hence, mass of water = 1500 x 1 = 1500 g

Gram molecular mass of water

= 2H + O

= (2 x 1) + 16 = 18 g

So, 18 g of water = 6.023 x 1023 molecules

Therefore, 1500 g of water will have

6.023×102318×1500=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}

Hence, the number of molecules in 1½ litres of water is 83.33 x 6.023 x 1023 molecules.

Question 10

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

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]

Hence, gram-atoms in 88.75 g of chlorine is 2.5 g. atoms

Question 11

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

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}

Hence, number of hydrogen atoms in 0.25 mole of H2SO4 = 0.5 x 6.023 x 1023 particles

Question 12

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

Answer

Gram molecules is the relative molecular mass of a substance expressed in grams.

Relative molecular mass of N2 = 2 x 14 = 28 g

Gram 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}

Hence, gram molecules in 21 g of nitrogen = 0.75 gram molecules

Question 13

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

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}

Hence, number of atoms in 10 litres of ammonia = 1.786 x 6.023 x 1023 atoms

Question 14

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

Answer

Gram atomic mass of Ne = 20 g

As, 1 mole of Ne weighs 20 g and has 6.023 x 1023 atoms

So, 60 g of Ne will have

6.023×102320×60=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}

Hence, number of atoms in 60 g of Ne = 3 x 6.023 x 1023 atoms

Question 15

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

Answer

Gram atomic mass of X (phosphorus) = 31 g

As, 31 g of phosphorus = 1 mole

So, 93 g of phosphorus = 131×93=3 moles\dfrac{1}{31} \times 93 = 3 \text{ moles}

Hence, number of moles in 93 g of X is 3 moles

Question 16

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

Answer

Gram molecular mass of O2 = 2 x 16 = 32 g

1 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.}

Volume occupied by 3.5 g of O2 gas at 27°C and 740 mm pressure:

s.t.p.given values
P1 = 760 mm of HgP2 = 740 mm of Hg
V1 = 2.45 litV2 = x lit
T1 = 273 KT2 = 27 + 273 K

Using the gas equation,

P1V1T1=P2V2T2\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}

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}

Hence, the volume occupied by 3.5 g of O2 gas at 27°C and 740 mm pressure is 2.76 lit.

Question 17

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

Answer

Gram molecular mass of sodium hydroxide
= Na + O + H
= 23 + 16 + 1
= 40 g

As, 40 g of sodium hydroxide = 1 mole

∴ 160 g of sodium hydroxide = 140×160=4 moles\dfrac{ 1}{40} \times 160 = 4 \text{ moles}

Hence, number of moles in 160 g of sodium hydroxide is 4

Question 18

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

Answer

Gram molecular mass of C2H6 = (2 x 12) + (6 x 1)
= 24 + 6
= 30 g

1 mole of C2H6 weighs 30 g

∴ 2.5 moles weighs = 301×2.5=75 g\dfrac{ 30}{1} \times 2.5 = 75 \text{ g}

Hence, weight of 2.5 moles of ethane [C2H6] is 75 g.

Question 19

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

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} \times 22.4 = 26 \text{ g}

As 22.4 lit of gas at s.t.p. weighs 26 g
∴ Molecular weight = 26 g

Question 20

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

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}

Hence, gram-atoms in 46 g of sodium is 2 g. atoms

Question 21

Calculate the number of moles of KClO3 that will be required to give 6 moles of oxygen.

Answer

2KClO3 ⟶ 2KCl + 3O2

O2:KClO33:26:x\begin{matrix} \text{O}_2 & : & \text{KClO}_3 & \\ 3 & : & 2 \\ 6 & : & x \end{matrix}

23×6=xx=4 moles\dfrac{2}{3} \times 6 = x \\[0.5em] \Rightarrow x = 4 \text{ moles}

Hence, 4 moles of KClO3 are required to give 6 moles of oxygen.

Question 22

Calculate the weight of the substance if it's molecular weight is 70 and in the gaseous form occupies 10 lits. at 27°C and 700 mm pressure.

Answer

Initial ConditionsFinal Conditions (s.t.p.)
P1 = 700 mm of HgP2 = 760 mm of Hg
V1 = 10 litV2 = x lit
T1 = 27 + 273 KT2 = 273 K

Using the gas equation,

P1V1T1=P2V2T2\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}

Substituting the values we get,

700×10300=760×x273x=700×10×273300×760x=1911228x=8.38 lit\dfrac{700 \times 10}{300} = \dfrac{760 \times x}{273} \\[0.5em] x = \dfrac{700 \times 10 \times 273}{300 \times 760 } \\[0.5em] x = \dfrac{1911}{228 } \\[0.5em] x = 8.38 \text{ lit}

1 gram molecular weight of the gas occupies 22.4 lit. at s.t.p.

∴ 70 g of the gas occupies 22.4 lit. at s.t.p.

If yy g of the gas occupies 8.38 lits, then

y=7022.4×8.38=26.18 gy = \dfrac{70}{22.4} \times 8.38 \\[0.5em] = 26.18 \text{ g}

Hence, weight of substance is 26.18 g

Question 23

State which has higher number of moles : 5 g. of N2O or 5 g. of NO [N = 14, O = 16]

Answer

Gram molecular mass of N2O
= (2 x 14) + 16
= 44 g

44 g of N2O = 1 mole

∴ 5 g of N2O = 144×5=0.11\dfrac{1}{44} \times 5 = 0.11 moles

Gram molecular mass of NO

= 14 + 16
= 30 g

30 g of NO = 1 mole

∴ 5 g of NO = 130×5=0.16\dfrac{1}{30} \times 5 = 0.16 moles

Hence, 5 g. of NO has higher moles than 5 g. of N2O

Question 24

State which has higher mass : 1 mole of CO2 or 1 mole of CO [C = 12, O = 16]

Answer

Gram molecular mass of CO2
= 12 + (2 x 16)
= 12 + 32
= 44 g

1 mole of CO2 = 44 g

Gram molecular mass of CO
= 12 + 16
= 28 g

1 mole of CO = 28 g

Therefore, 1 mole of CO2 has higher mass.

Question 25

State which has higher no. of atoms : 1 g of O2 or 1 g of Cl2 [O = 16, Cl = 35.5]

Answer

Molecular mass of O2 = 2 x 16 = 32 g

32 g of O2 = 2 x 6.023 x 1023 atoms

∴ 1 g of O2

=2×6.023×102332=0.0625×6.023×1023 atoms= \dfrac{2 \times 6.023 \times 10^{23}}{32} \\[0.5em] = 0.0625 \times 6.023 \times 10^{23} \text{ atoms}

Molecular mass of Cl2 = 2 x 35.5 = 71 g

71 g of Cl2 = 2 x 6.023 x 1023 atoms

∴ 1 g of Cl2

=2×6.023×102371=0.0281×6.023×1023 atoms= \dfrac{2 \times 6.023 \times 10^{23}}{71} \\[0.5em] = 0.0281 \times 6.023 \times 10^{23} \text{ atoms}

Hence, 1 g of O2 has more number of atoms.

Vapour Density And Molecular Weight

Question 1

500 ml. of a gas 'X' at s.t.p. weighs 0.50 g. Calculate the vapour density and molecular weight of the gas. [1 lit. of H2 at s.t.p. weighs 0.09 g].

Answer

Given, 500 ml. of gas 'X' at s.t.p. weighs 0.50 g

Therefore, 1000 ml of gas 'X' at s.t.p. will weigh (0.50 x 2) g

Vapour density of gas X =

Wt. of 1000 ml of gas at s.t.p.Wt. of 1000 ml of H2 at s.t.p.=0.50×20.09=11.1\dfrac{\text{Wt. of 1000 ml of gas at s.t.p.}}{\text{Wt. of 1000 ml of H}_2 \text{ at s.t.p.}} \\[0.5em] = \dfrac{0.50 \times 2}{0.09} \\[0.5em] = 11.1

Hence, vapour density of gas is 11.11

Molecular weight = 2 x Vapour density
= 2 x 11.1
= 22.2 g

Hence, molecular weight of gas is 22.2 g

Question 2

A gas cylinder holds 85 g of a gas 'X'. The same cylinder when filled with hydrogen holds 8.5 g of hydrogen under the same conditions of temperature and pressure. Calculate the molecular weight of 'X'.

Answer

Vapour density of gas X =

Wt. of certain volume of gas Wt. of same volume of H2=858.5=10\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \dfrac{85}{8.5} \\[0.5em] = 10

Molecular weight = 2 x Vapour density
= 2 x 10 = 20 g

Hence, molecular weight of gas X is 20 g

Question 3

Calculate the relative molecular mass [molecular weight] of 290 ml. of a gas 'A' at 17 °C and 1520 mm pressure which weighs 2.73 g at s.t.p. [1 litre of hydrogen at s.t.p. weighs 0.09 g.]

Answer

Convert the volume to s.t.p. using gas equation

Initial ConditionsFinal Conditions (s.t.p.)
P1 = 1520 mm of HgP2 = 760 mm of Hg
V1 = 290 mlV2 = x lit
T1 = 17 + 273 KT2 = 273 K

Using the gas equation,

P1V1T1=P2V2T2\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}

Substituting the values we get,

1520×290290=760×x273x=1520×273760x=546 ml\dfrac{1520 \times 290}{290} = \dfrac{760 \times x}{273} \\[0.5em] x = \dfrac{1520 \times 273}{760} \\[0.5em] x = 546 \text{ ml}

546 ml of gas at s.t.p. weighs 2.73 g

∴ Wt. of 1000 ml of gas = 2.73546×1000=5 g\dfrac{2.73}{546} \times 1000 = 5 \text { g}

Vapour density of gas X =

Wt. of 1 lit of gas XWt. of 1 lit of H2=50.09=55.55\dfrac{\text{Wt. of 1 lit of gas X}}{\text {Wt. of 1 lit of H}_2} \\[0.5em] = \dfrac{5}{0.09} \\[0.5em] = 55.55

Hence, vapour density of gas is 55.55

Molecular weight = 2 x Vapour Density
= 2 x 55.555
= 111.11 g

Hence, molecular weight of gas is 111.11 g

Question 4

State the volume occupied by 40 g of a hydrocarbon – CH4 at s.t.p. if it's V.D. is 8.

Answer

Vapour density of hydrocarbon = 8

∴ Molecular weight = 2 x Vapour density
= 2 x 8 = 16 g

16 g of hydrocarbon at s.t.p. occupy 22.4 lit

∴ 40 g of hydrocarbon will occupy = 22.416×40=56 lit.\dfrac{22.4}{16} \times 40 = 56 \text{ lit.}

Hence, volume occupied by 40 g of a hydrocarbon (CH4) at s.t.p. is 56 lit.

Question 5

Calculate the atomicity of a gas X [at. no. 35.5] whose vapour density is equal to it's relative atomic mass.

Answer

Given,

Vapour density = relative atomic mass = 35.5

Molecular weight = 2 x Vapour Density = 2 x 35.5 = 71 g

Atomicity is the number of atoms present in one molecule of that element.

Number of atoms

=Molecular weightAtomic weight=7135.5=2= \dfrac{\text{Molecular weight}}{\text{Atomic weight}} \\[0.5em] = \dfrac{\text{71}}{\text{35.5}} \\[0.5em] = 2

Hence, atomicity of a gas X is 2

Question 6

Calculate the relative molecular mass and vapour density of methyl alcohol [CH3OH] if 160 g. of the alcohol on vaporization has a volume of 112 litres at s.t.p.

Answer

Given,

Weight of 112 lit of CH3OH = 160 g [at s.t.p.]

∴ Weight of 1 lit of CH3OH = 160112\dfrac{160}{112} = 1.4286 g

Weight of 1 lit of H2 = 0.09 g [at s.t.p.]

Under similar temperature and pressure,

Vapour density of gas methyl alcohol =

Wt. of certain vol. of gasWt. of same vol. of H2=1.42860.09=15.8716\dfrac{\text{Wt. of certain vol. of gas}}{\text {Wt. of same vol. of H}_2} \\[0.5em] = \dfrac{1.4286}{0.09} \\[0.5em] = 15.87 \approx 16

Molecular weight = 2 x Vapour density
= 2 x 15.87 = 31.74 g ≈ 32 g

Hence, molecular weight of methyl alcohol is 32 g and vapour density of methyl alcohol is 16

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