Gay Lussac's Law

[By Lussac's law]

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

To calculate the volume of ammonia gas formed.

$$\begin{matrix} \text{H}_2 & : & \text{NH}_3 \\ 3 & : & 2 \\ 6 & : & x \end{matrix}$$

Therefore,

$$\begin{aligned}\frac{2}{3} \times 6 = x \\[0.5em] \Rightarrow x = 4 \text{ lts}\end{aligned}$$

Hence, volume of ammonia gas formed is 4 lts

[By Lussac's law]

$$\begin{matrix} \text{\scriptsize{2C}}_2\text{\scriptsize{H}}_6 & \text{\scriptsize{+}} & \text{\scriptsize{7O}}_2 & \longrightarrow & \text{\scriptsize{4CO}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{6H}}_2\text{\scriptsize{O}} \\ \text{\scriptsize{2 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{7 vol.}} & \longrightarrow & \text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{6 vol.}} \end{matrix}$$

To calculate the volume of unused oxygen.

$$\begin{matrix} \text{C}_2\text{H}_6 & : &\text{O}_2 \\ 2 & : & 7 \\ 600 & : & x \end{matrix}$$

Therefore,

$$\begin{aligned}\frac{7}{2} \times 600 = x \\[0.5em] \Rightarrow x = 2100 \text{cc}\end{aligned}$$

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

To calculate the volume of carbon dioxide formed

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

Therefore,

$$\begin{aligned}\frac{4}{2} \times 600 = x \\[0.5em] \Rightarrow x = 1200 \text{cc}\end{aligned}$$

Therefore, volume of carbon dioxide formed is 1200 cc

Given,

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

[By Lussac's law]

$$\begin{matrix} 2\text{CO}& + & \text{O}_2 & \longrightarrow & 2\text{CO}_2 & \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} & \\ \end{matrix}$$

To calculate the amount of CO₂ produced,

$$\begin{matrix} \text{CO}& : & \text{CO}_2 & \\ 2 \text{ vol.} & : & 2 \text{ vol.} & \\ 10 \text{ ml} & : & x \text{ ml} & \\ \end{matrix}$$

Therefore, CO₂ produced is 10 ml

To calculate the amount of O₂ used,

$$\begin{matrix} \text{CO}& : & \text{O}_2 & \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \\ 10 \text{ ml} & : & x \text{ ml} & \\ \end{matrix}$$

$$\begin{aligned}\frac{1}{2} \times 10 = x \\[0.5em] \Rightarrow x = 5 \text{ ml}\end{aligned}$$

From relation,

$$\begin{matrix} 2\text{H}_2 & + & \text{O}_2 & \longrightarrow & 2\text{H}_2 \text {O} & \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} & \\ \end{matrix}$$

$$\begin{matrix} \text{H}_2 & : & \text{O}_2 & \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \\ 20 \text{ ml} & : & x \text{ ml} & \\ \end{matrix}$$

$$\begin{aligned}\frac{1}{2} \times 20 = x \\[0.5em] \Rightarrow x = 10 \text{ ml}\end{aligned}$$

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.

[By Lussac's law]

$$\begin{matrix} \text{\scriptsize{4NH}}_3 & \text{\scriptsize{+}} & \text{\scriptsize{5O}}_2 & \longrightarrow & \text{\scriptsize{4NO}} & \text{\scriptsize{+}} & \text{\scriptsize{6H}}_2\text{\scriptsize{O}} \\ \text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{5 vol.}} & \longrightarrow &\text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{6 vol.}} \end{matrix}$$

To calculate the volume of oxygen required.

$$\begin{matrix} \text{NH}_3 & : & \text{O}_2 & \\ 4 & : & 5 \\ 224 & : & x \end{matrix}$$

Therefore,

$$\begin{aligned}\frac{5}{4} \times 224 = x \\[0.5em] \Rightarrow x = 280 \text{ cm}^3\end{aligned}$$

Therefore, volume of oxygen required is 280 cm³.

[By Lussac's law]

$$\begin{matrix} \text{\scriptsize{2C}}_2 \text{\scriptsize{H}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{5O}}_2 & \longrightarrow & \text{\scriptsize{4CO}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{2H}}_2\text{\scriptsize{O}} \\ \text{\scriptsize{2 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{5 vol.}} & \longrightarrow & \text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{2 vol.}} \end{matrix}$$

To calculate the volume of air required,

$$\begin{matrix} \text{C}_2 \text{H}_2 & : & \text{O}_2 & \\ 2 & : & 5 \\ 50 & : & x \end{matrix}$$

Therefore, volume of oxygen (x),

$$\begin{aligned}\frac{5}{2} \times 50 = x \\[0.5em] \Rightarrow x = 125 \text{ cm}^3\end{aligned}$$

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

$$\begin{aligned}\frac{100}{20} \times 125 = x \\[0.5em] \Rightarrow x = 625 \text{ cm}^3\end{aligned}$$

Therefore, volume of air required is 625 cm³.

[By Lussac's law]

$$\begin{matrix} \text{\scriptsize{2C}}_2 \text{\scriptsize{H}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{5O}}_2 & \longrightarrow & \text{\scriptsize{4CO}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{2H}}_2\text{\scriptsize{O}} \\ \text{\scriptsize{2 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{5 vol.}} & \longrightarrow & \text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{2 vol.}} \end{matrix}$$

To calculate the volume of acetylene :

$$\begin{matrix} \text{CO}_2 & : & \text{C}_2 \text{H}_2 & \\ 4 & : & 2 \\ 200 & : & x \end{matrix}$$

Therefore,

$$\begin{aligned}\frac{2}{4} \times 200 = x \\[0.5em] \Rightarrow x = 100 \text { cm}^3\end{aligned}$$

Hence, volume of acetylene in the original mixture is 100 cm³.

To calculate the volume of oxygen :

$$\begin{matrix} \text{CO}_2 & : & \text{O}_2 & \\ 4 & : & 5 \\ 200 & : & x \end{matrix}$$

Therefore,

$$\begin{aligned}\frac{5}{4} \times 200 = x \\[0.5em] \Rightarrow x = 250 \text { cm}^3\end{aligned}$$

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

[By Lussac's law]

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

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

$\therefore 10\% \text{ of 2 lit} = \frac{10}{100} \times 2 = 0.2 \text{ lit}$

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

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]

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

To calculate the amount of O₂ used,

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

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,

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

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

[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

$= \frac{44}{22.4} \times 2.8 = 5.5 \text{ g}$

Therefore, mass of 2.8 litres of CO₂ is 5.5 g

[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

$= \frac{22.4}{71} \times 53.5 = 16.87 \text{ lit}$

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

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

$$\begin{aligned}= \frac{6.023 \times 10^{23}}{36.5} \times 109.5 \\[0.5em] = \frac{109.5}{36.5} \times 6.023 \times 10^{23} \\[0.5em] = 3 \times 6.023 \times 10^{23} \\[0.5em]\end{aligned}$$

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

(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

$$\begin{aligned}= \frac{6.023 \times 10^{23}}{256} \times 192 \\[0.5em] = \frac{192}{256} \times 6.023 \times 10^{23} \\[0.5em] = 0.75 \times 6.023 \times 10^{23} \text { molecules} \\[0.5em]\end{aligned}$$

(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

$$\begin{aligned}= \frac{6.023 \times 10^{23}}{32} \times 192 \\[0.5em] = \frac{192}{32} \times 6.023 \times 10^{23} \\[0.5em] = 6 \times 6.023 \times 10^{23} \text { atoms} \\[0.5em]\end{aligned}$$

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

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

$$\begin{aligned}= \frac{6.023 \times 10^{23}}{39} \times 117 \\[0.5em] = \frac{117}{39} \times 6.023 \times 10^{23} \\[0.5em] = 3 \times 6.023 \times 10^{23} \\[0.5em]\end{aligned}$$

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

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

$$\begin{aligned}= \frac{1}{331} \times 19.86 = 0.06 \text { moles} \\[0.5em]\end{aligned}$$

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

$$\begin{aligned}= \frac{ 6.023 \times 10^{23}}{331} \times 19.86 \\[0.5em] = \frac{19.86}{331} \times 6.023 \times 10^{23} \\[0.5em] = 0.06 \times 6.023 \times 10^{23} \text{molecules}\end{aligned}$$

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

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

$= \frac{202}{6.023 \times 10^{23}} = 33.53 \times 10^{-23} \text{g}$

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

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

$$\begin{aligned}\frac{6.023 \times 10^{23}}{18} \times 1500 \\[0.5em] = \frac{1500}{18} \times 6.023 \times 10^{23} \\[0.5em] = 83.33 \times 6.023 \times 10^{23}\end{aligned}$$

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

Gram atoms is the relative atomic mass of an element expressed in grams.

$$\begin{aligned}\text{Gram atoms} = \frac{\text{Mass in grams}}{\text{Relative At. mass [At. wt.]}} \\[0.5em] = \frac{88.75}{35.5} \\[0.5em] = 2.5 \text { gram atoms} \\[0.5em]\end{aligned}$$

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

1 mole of hydrogen atom has 2 x 6.023 x 10²³ atoms.

∴ 0.25 mole will have

$$\begin{aligned}= \frac{2 \times 6.023 \times 10^{23}}{1} \times 0.25 \\[0.5em] = 0.5 \times 6.023 \times 10^{23} \text{ atoms}\end{aligned}$$

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

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

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

$$\begin{aligned}\text{Gram molecules} = \frac{\text{Mass in grams [of nitrogen]}}{\text{Rel. molecular mass [Mol. wt.]}} \\[0.5em] = \frac{21}{28} \\[0.5em] = 0.75 \text { gram molecules}\end{aligned}$$

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

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

$$\begin{aligned}= \frac{4 \times 6.023 \times 10^{23}}{22.4} \times 10 \\[0.5em] = \frac{40}{22.4} \times 6.023 \times 10^{23} \\[0.5em] = 1.786 \times 6.023 \times 10^{23} \text{ atoms}\end{aligned}$$

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

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

$$\begin{aligned}\frac{ 6.023 \times 10^{23}}{20} \times 60 \\[0.5em] = \frac{60}{20} \times 6.023 \times 10^{23} \\[0.5em] = 3 \times 6.023 \times 10^{23} \text{ atoms}\end{aligned}$$

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

Gram atomic mass of X (phosphorus) = 31 g

As, 31 g of phosphorus = 1 mole

So, 93 g of phosphorus = $\frac{1}{31} \times 93 = 3 \text{ moles}$

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

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 = $\frac{22.4}{32} \times 3.5 = 2.45 \text{ lit.}$

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 Hg	P2 = 740 mm of Hg
V1 = 2.45 lit	V2 = x lit
T1 = 273 K	T2 = 27 + 273 K

Using the gas equation,

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substituting the values we get,

$$\begin{aligned}\frac{760 \times 2.45}{273} = \frac{740 \times x}{300} \\[0.5em] x = \frac{760 \times 2.45 \times 300}{740 \times 273 } \\[0.5em] x = \frac{5,58,600}{2,02,020} \\ \\[0.5em] x = 2.76 \text{ lit}\end{aligned}$$

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

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 = $\frac{ 1}{40} \times 160 = 4 \text{ moles}$

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

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 = $\frac{ 30}{1} \times 2.5 = 75 \text{ g}$

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

Given,

2.24 lit volume of gas weighs 2.6 g

∴ 22.4 lit volume of gas weighs = $\frac{2.6}{2.24} \times 22.4 = 26 \text{ g}$

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

Gram atoms is the relative atomic mass of an element expressed in grams.

$$\begin{aligned}\text{Gram atoms} = \frac{\text{Mass in grams [of sodium]}}{\text{Rel. atomic mass [At. wt.]}} \\[0.5em] = \frac{46}{23} = 2 \text { gram atoms}\end{aligned}$$

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

2KClO₃ ⟶ 2KCl + 3O₂

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

$$\begin{aligned}\frac{2}{3} \times 6 = x \\[0.5em] \Rightarrow x = 4 \text{ moles}\end{aligned}$$

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

Initial Conditions	Final Conditions (s.t.p.)
P1 = 700 mm of Hg	P2 = 760 mm of Hg
V1 = 10 lit	V2 = x lit
T1 = 27 + 273 K	T2 = 273 K

Using the gas equation,

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substituting the values we get,

$$\begin{aligned}\frac{700 \times 10}{300} = \frac{760 \times x}{273} \\[0.5em] x = \frac{700 \times 10 \times 273}{300 \times 760 } \\[0.5em] x = \frac{1911}{228 } \\[0.5em] x = 8.38 \text{ lit}\end{aligned}$$

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 $y$ g of the gas occupies 8.38 lits, then

$$\begin{aligned}y = \frac{70}{22.4} \times 8.38 \\[0.5em] = 26.18 \text{ g}\end{aligned}$$

Hence, weight of substance is 26.18 g

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

44 g of N₂O = 1 mole

∴ 5 g of N₂O = $\frac{1}{44} \times 5 = 0.11$ moles

Gram molecular mass of NO

= 14 + 16
= 30 g

30 g of NO = 1 mole

∴ 5 g of NO = $\frac{1}{30} \times 5 = 0.16$ moles

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

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.

Molecular mass of O₂ = 2 x 16 = 32 g

32 g of O₂ = 2 x 6.023 x 10²³ atoms

∴ 1 g of O₂

$$\begin{aligned}= \frac{2 \times 6.023 \times 10^{23}}{32} \\[0.5em] = 0.0625 \times 6.023 \times 10^{23} \text{ atoms}\end{aligned}$$

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

$$\begin{aligned}= \frac{2 \times 6.023 \times 10^{23}}{71} \\[0.5em] = 0.0281 \times 6.023 \times 10^{23} \text{ atoms}\end{aligned}$$

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

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 =

$$\begin{aligned}\frac{\text{Wt. of 1000 ml of gas at s.t.p.}}{\text{Wt. of 1000 ml of H}_2 \text{ at s.t.p.}} \\[0.5em] = \frac{0.50 \times 2}{0.09} \\[0.5em] = 11.1\end{aligned}$$

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

Vapour density of gas X =

$$\begin{aligned}\frac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \frac{85}{8.5} \\[0.5em] = 10\end{aligned}$$

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

Hence, molecular weight of gas X is 20 g

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

Initial Conditions	Final Conditions (s.t.p.)
P1 = 1520 mm of Hg	P2 = 760 mm of Hg
V1 = 290 ml	V2 = x lit
T1 = 17 + 273 K	T2 = 273 K

Using the gas equation,

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substituting the values we get,

$$\begin{aligned}\frac{1520 \times 290}{290} = \frac{760 \times x}{273} \\[0.5em] x = \frac{1520 \times 273}{760} \\[0.5em] x = 546 \text{ ml}\end{aligned}$$

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

∴ Wt. of 1000 ml of gas = $\frac{2.73}{546} \times 1000 = 5 \text { g}$

Vapour density of gas X =

$$\begin{aligned}\frac{\text{Wt. of 1 lit of gas X}}{\text {Wt. of 1 lit of H}_2} \\[0.5em] = \frac{5}{0.09} \\[0.5em] = 55.55\end{aligned}$$

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

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 = $\frac{22.4}{16} \times 40 = 56 \text{ lit.}$

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

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

$$\begin{aligned}= \frac{\text{Molecular weight}}{\text{Atomic weight}} \\[0.5em] = \frac{\text{71}}{\text{35.5}} \\[0.5em] = 2\end{aligned}$$

Hence, atomicity of a gas X is 2

Given,

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

∴ Weight of 1 lit of CH₃OH = $\frac{160}{112}$ = 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 =

$$\begin{aligned}\frac{\text{Wt. of certain vol. of gas}}{\text {Wt. of same vol. of H}_2} \\[0.5em] = \frac{1.4286}{0.09} \\[0.5em] = 15.87 \approx 16\end{aligned}$$

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