Chemistry Test - 6

Here \(\mathrm{CO}_{3}^{2-}\) and \(\mathrm{H}_{2} \mathrm{O}\) are Bronsted bases as water accept \(\mathrm{H}^{+}\) and \(\mathrm{CO}_{3}^{2-}\) is the conjugate base of \(\mathrm{HCO}_{3}^{-}\) and as the reaction is reversible so \(\mathrm{CO}_{3}^{2-}\) is accepting \(\mathrm{H}^{+}\).

\(\mathrm{HCO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CO}_{3}^{2-}+\mathrm{H}_{3} \mathrm{O}^{+}\)

Number of moles of \(\mathrm{CH}_{4}(\mathrm{~g})\) used \(=\frac{1 \mathrm{~atm} \times 2 \mathrm L}{.0821 \mathrm{~L} \mathrm{~atm~K^{-1 } \mathrm { mol }}{ }^{-1} \times 573 \mathrm{~K}}=0.0425 \mathrm{~mol}\)

Number of moles of \(\mathrm{O}_{2}(\mathrm{~g})\) taken \(=\frac{4 \mathrm{~atm} \times 2 \mathrm{~L}}{.0821 \mathrm{~L} \mathrm{~atm~K} \mathrm{L}^{-1} \mathrm{~mol}^{-1} \times 573 \mathrm{~K}}=0.1700 \mathrm{~mol}\)

Here, methane is the limiting reactant. Thus, according to the balanced equation, the number of moles of \(\mathrm{CO}_{2}\) formed is the same as the number of moles of methane reacted i.e. \(0.0525 \mathrm{~mol}\)

So, mass of \(\mathrm{CO}_{2}(\mathrm{~g})\) formed \(=0.0425 \mathrm{~mol} \times 44 \mathrm{~g} \mathrm{~mol}^{-1}=1.87 \mathrm{~g}\)

Higher stability of allyl and benzyl carbocations is due to dispersal of positive charge by resonance

whereas in alkyl carbocations dispersal of positive charge on different hydrogen atoms is due to inductive effect. Hence the correct order of stability will be

Aldehydes with no \(\alpha\) - \(H\) atom undergo Cannizzaro reaction on Heating with conc. alkali solution.

Of all the given compounds, only \(\mathrm{C _6 H _5 CHO}\) has no \(\alpha\)-hydrogen. So, it will undergo Cannizzaro reaction.

The three dimensional arrangement of constituent particles in a crystal is represented in such a way each particle is taken as a point, the arrangement is called as crystal lattice.

Thus, a regular arrangement of the points in space is the correct definition of crystal lattice.

Compound \(\mathrm{CCl}_{3} \mathrm{CH}(\mathrm{OH})_{2}\) is stable.

Here, we noticed that two OH group attached with one carbon atom at all three compounds which is called the geminal diol compound.

(C) \(\mathrm{CCl}_{3} \mathrm{CH}(\mathrm{OH})_{2}\)

(A) \(\mathrm{CH}_{3}-\mathrm{CH}(\mathrm{OH})_{2}\)

(B) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH})_{2}\)

Dehydration reaction:

(C) \( \mathrm{CCL}_{3} \mathrm{CH}(\mathrm{OH})_{2}\underset{\mathrm{H^+}}{\stackrel{\mathrm{C}}{\longrightarrow}} \mathrm{Cl}_{3} \mathrm{CH}\left(\mathrm{OH}_{2}\right)-\mathrm{O}-\mathrm{H} \underset{-\mathrm{H}_{2} \mathrm{O}}{\stackrel{\mathrm{H^+}}{\longrightarrow}} \mathrm{CH}_{3}-\mathrm{C}-\mathrm{CHO}\)

(A) \( \mathrm{CH}_{3}-\mathrm{CH}(\mathrm{OH})_{2} \stackrel{\mathrm{H^+}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{OH}_{2}^{+}\right)-\mathrm{O}-\mathrm{H} \underset{-\mathrm{H}_{2} \mathrm{O}}{\stackrel{\mathrm{H^+}}{\longrightarrow}} \mathrm{CH}_{3}-\mathrm{CHO}\)

(B) \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH})_{2} \stackrel{\mathrm{H}^{+}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) 2 \mathrm{C}\left(\mathrm{OH}_{2}\right)-\mathrm{O}-\mathrm{H} \underset{-\mathrm{H}_{2} \mathrm{O}}{\stackrel{\mathrm{H^+}}{\longrightarrow}} \mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{CHO}\)

\(\because+\) I group (Alkyl group) : geminal diol compound stability decreases.

\(\because-\) I group (Halogen group) : geminal diol compound stability increases.

Hence the correct option is (C).

2KI + Cl₂ → 2KCl l₂

I₂ CCl₄ → Violet Colour

But the excess of Cl₂ should be avoided.

The layer may become colourless due to conversion of I₂ to HIO₃

I₂ + 5Cl₂ + 6H₂O → 2HIO₃ + 10HCl

In case of Br₂

Br₂ + 2H₂O + Cl₂ → 2HBrO + HCl

In surface chemistry,

1. Dissipation/ Desorption is the process of removing an adsorbed substance from a surface on which it is absorbed.

2. Absorption is a process in which the substance (adsorbate) is uniformly distributed throughout the bulk.

3. Sorption is a process where both the phenomenon of adsorption and absorption take place simultaneously.

4. Physisorption is a type of adsorption where weak Van der Waals forces act between the adsorbate and adsorbent.

Thus, the correct combination of an answer will be A-2, B-3, C-4, D-1

∵ ΔG = ΔH − TΔS = −ve

Given: ΔS = −ve

ΔH = −ve

∴ To get ΔG = −ve

ΔS must be less then ΔH i.e. ΔH > TΔS

Thus, the reaction is exotherms and favours and increases with the decrease in temperature.

Given that:

The solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass.

Therefore,

Total amount of solute present in the mixture will be given by,

\(300 \times \frac{25}{100}+400 \times \frac{40}{100}\)

\(=75+160\)

\(=235 \mathrm{~g}\)

Total amount of solution \(=300+400=700 \mathrm{~g}\)

Therefore, 

Mass percentage of the solute in the resulting solution \(=\frac{\text{Total amount of solute}}{\text{Total amount of solution}} \times100\)

\(=\frac{235}{700} \times100\)

\(=33.57 \%\)

Then, mass percentage of the solvent in the resulting solution will be:

\(=(100-33.57) \%\)

\(=66.43 \%\)