NEET and AIPMT NEET Chemistry States of Matter MCQ Questions Solved

(d) Given, number of moles of hydrogen (${\mathrm{n}}_{{\mathrm{H}}_2}$) and that of oxygen (${\mathrm{n}}_{{\mathrm{O}}_2}$) are equal.

.^.. We have, the relation between ratio of number of moles escaped and ratio of molecular mass.

                 $\frac{{\mathrm{n}}_{{\mathrm{H}}_2}}{{\mathrm{n}}_{{\mathrm{O}}_2}}=\sqrt{\frac{{\mathrm{M}}_{{\mathrm{H}}_2}}{{\mathrm{M}}_{{\mathrm{O}}_2}}}$

where, M = Molecular mass of the molecule

               $\frac{{\mathrm{n}}_{{\mathrm{O}}_2}}{{\mathrm{n}}_{{\mathrm{H}}_2}}=\sqrt{\frac{2}{32}}=\sqrt{\frac{1}{16}}$

                $\frac{{\mathrm{n}}_{{\mathrm{O}}_2}}{0.5}$ = $\frac{1}{4}$

                 ${\mathrm{n}}_{{\mathrm{O}}_2}$= 0.5/4 =1/8

Real gases show ideal gas behaviour at high temperature and low pressures.

(c) According to Avogadro's hypothesis, Volume of a gas (V) $\propto$ number of moles (n)

Therefore, the ratio of the volumes of gases can be determined in terms of their moles. The ratio of volumes of H₂: O₂:methane (CH₄)
is given by

$$\begin{gathered} {\mathrm{V}}_{{\mathrm{H}}_2} :\hspace{0.17em}{\mathrm{V}}_{{\mathrm{O}}_2}: {\mathrm{V}}_{{\mathrm{CH}}_4} = {\mathrm{n}}_{{\mathrm{H}}_2} :\hspace{0.17em}{\mathrm{n}}_{{\mathrm{O}}_2} : {\mathrm{n}}_{{\mathrm{CH}}_4} \\ \Rightarrow {\mathrm{V}}_{{\mathrm{H}}_2} :\hspace{0.17em}{\mathrm{V}}_{{\mathrm{O}}_2}: {\mathrm{V}}_{{\mathrm{CH}}_4} = \frac{{\mathrm{m}}_{{\mathrm{H}}_2}}{{\mathrm{M}}_{{\mathrm{H}}_2}} : \frac{{\mathrm{m}}_{{\mathrm{O}}_2}}{{\mathrm{M}}_{{\mathrm{O}}_2}} : \frac{{\mathrm{m}}_{{\mathrm{CH}}_4}}{{\mathrm{M}}_{{\mathrm{CH}}_4}} \\ \mathrm{But} {\mathrm{m}}_{{\mathrm{H}}_2} = {\mathrm{m}}_{{\mathrm{O}}_2} = {\mathrm{m}}_{{\mathrm{CH}}_4} = \frac{\mathrm{m}}{2} :\hspace{0.17em}\frac{\mathrm{m}}{32}: \frac{\mathrm{m}}{16} = 16:1:2 \end{gathered}$$

The given problem is related to the concept of stiochiometry of chemical equations. Thus, we have to convert the given volumes into their moles and then, identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction].

The limiting reagent gives the moles of product formed in the reaction.

              H₂(g) + Cl₂(g) $\to$2HCl (g)

Initial vol   22.4 L    11.2 L      2 mol

... 22.4 L volume at STP is occupied by,

              Cl₂ = 1 mole,

... 11.2 L volume will be occupied by,

             Cl₂ = 1x 11.2/22.4 mole = 0.5 mol

Thus, H₂(g) + Cl₂(g) $\to$2HCl (g)

(d) Easily liquefiable gases like NH₃, SO₂ etc. exhibit maximum deviation from ideal gas as for them Z<<< 1 .

${\mathrm{CH}}_4$  also exhibits deviation but it is less as compared to NH₃.

"CO="N₂ 

P_co = P_N2 

Given, P_co + P_N2 = 1atm 

or       2P_N2 = 1 atm

or        P_N2 = 0.5atm  

(a)

 $$\begin{gathered} \frac{{\mathrm{V}}_{\mathrm{A}}}{{\mathrm{V}}_{\mathrm{B}}} = \sqrt{\frac{{\mathrm{M}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{A}}}} \\ \mathrm{or} \frac{{\mathrm{V}}_{\mathrm{A}}}{{{\displaystyle \mathrm{t}}}_{\mathrm{A}}} \times \frac{{\mathrm{t}}_{\mathrm{B}}}{{\mathrm{V}}_{\mathrm{B}}} = \sqrt{\frac{{\displaystyle {\mathrm{M}}_{\mathrm{B}}}}{{\displaystyle {\mathrm{M}}_{\mathrm{A}}}}} \\ \mathrm{or} \frac{10}{20} = \sqrt{\frac{{\displaystyle {\mathrm{M}}_{\mathrm{B}}}}{{\displaystyle 49}}} \\ \mathrm{or} \frac{1}{4} = \frac{{\mathrm{M}}_{\mathrm{a}}}{49} \\ \therefore {\mathrm{M}}_{\mathrm{B}} = \frac{49}{4} = 12.25 \mathrm{u}. \end{gathered}$$

Average velocity = $\sqrt{\frac{8\mathrm{RT}}{\mathrm{\pi M}}}$

V_av $\mathrm{\alpha }$ $\sqrt{\mathrm{T}}$

or (V_av)₂ / (V_av )₁ = $\sqrt{\frac{2\mathrm{T}}{\mathrm{T}}}$ = 1.4

Key Idea: KE = 3/2RT

KE ∝ T

Here, the temperature of gas remains constant, hence kinetic energy of molecules remains the same.

Key Idea: Volume of a gas STP = 22.4L

To burn 22.4 L C₃H₈ the oxygen required is = 5 x 22.4 L

To burn 1 LC₃H₈ the oxygen required will be = 5 x 22.4 / 22.4 = 5L