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Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight Chemistry Solutions

When 9.45g of ClCH₂COOH is added to 500ml. of water, its freezing point drops by 0.5°C. The dissociation constant of ClCH₂ COOH is x * 10^-3. The value of x is__________ (Rounded off to the nearest integer)

$[k_{f (H_{2} O)} = 1.86 K k g m o l^{- 1}]$

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alok kumar singh

Contributor-Level 10

$Δ T_{f} = 0.5 ° C$

K_f = 1.86

Using, density of water = 1g / mL

i = ?

$Δ T_{f} = i k_{f} . m$

$0.5 = i * 1.86 * \frac{9.45}{94.5 * 500} * 1000$

i = 1.344

Now, using $α = \frac{i - 1}{n - 1}$

n for ClCH₂COOH = 2

a = $\frac{1.344 - 1}{2 - 1} = 0.344$

Using

$K_{a} = \frac{C α^{2}}{1 - α}$

$K_{a} = \frac{0.2 * {(0.344)}^{3}}{0.66} = 36 * 1 0^{- 3}$

So; x = 36

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight Equilibrium

At 1990K and 1 atm pressure, there are equal number of Cl₂ molecules and Cl atoms in the reaction mixture. The value of K_P for the reaction $?$ Cl_2(g) 2Cl_(g) under the above conditions is x * 10^-1. The value of x is______ (Rounded off to the nearest integer)

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alok kumar singh

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Initially -> 1 mol -

At eq. 1-x mol 2x mol

Here; molecules of Cl₂ = atoms of Cl

i.e moles of Cl₂ = moles of Cl

So : 1 – x = 2x x = 1/3

Moles of Cl₂ at equibrium = $\frac{2}{3}$

Moles of Cl at equilibrium = $\frac{2}{3}$

Total moles = $\frac{4}{3}$

...more

Initially -> 1 mol -

At eq. 1-x mol 2x mol

Here; molecules of Cl₂ = atoms of Cl

i.e moles of Cl₂ = moles of Cl

So : 1 – x = 2x x = 1/3

Moles of Cl₂ at equibrium = $\frac{2}{3}$

Moles of Cl at equilibrium = $\frac{2}{3}$

Total moles = $\frac{4}{3}$

Now: $P_{C l_{2}} = \frac{2}{4} * 1 a t m$

$= \frac{1}{2} a t m$

$P_{C l} = \frac{2}{4} * 1 a t m = \frac{1}{2} a t m$

$K_{P} = \frac{{(P_{C l})}^{2}}{P_{C l_{2}}} = \frac{{(\frac{1}{2})}^{2}}{\frac{1}{2}} = \frac{1}{2}$

$K_{P} = 0.5 = 5 * 1 0^{- 1}$

Here : x = 5

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight Equilibrium

For the reaction A_(g) ® B_(g) the value of the equilibrium constant at 300K and 1 atm is equal to 100.0 The value of for the reaction at 300K and 1 atm in J mol^-¹ is -xR, where x is_________.

(Rounded off to the nearest integer)

[R = 8.31 J mol^-¹ K^-¹ and In 10 = 2.3]

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alok kumar singh

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A (g) -> B (g) K_P = 100

$Δ G$ at 300 K and 1 atm

Using

$Δ G = - R T l n K_{P}$

$Δ G = - R * 300 l n 100$

=-R * 300 * 2 * 2.3

$Δ G = - 1380 R$

So; x = 1380.

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight Chemistry Coordination Compounds

The stepwise formation of ${[C u {(N H_{3})}_{4}]}^{2}^{+}$ is given below:

$C u^{2 +} + N H_{3} \overset{k_{1}}{?} {[C u (N H_{3})]}^{2 +}$

${[C u (N H_{3})]}^{2 +} + N H_{3} \overset{k_{2}}{?} {[C u {(N H_{3})}_{2}]}^{2 +}$

${[C u {(N H_{3})}_{2}]}^{2 +} + N H_{3} \overset{k_{3}}{?} {[C u {(N H_{3})}_{3}]}^{2 +}$

${[C u {(N H_{3})}_{3}]}^{2 +} + N H_{3} \overset{k_{4}}{?} {[C u {(N H_{3})}_{4}]}^{2 +}$

The value of stability constants k₁, k₂, k₃ and k₄ are $1 0^{4}, 1.58 * 1 0^{3}, 5 * 1 0^{2} a n d 1 0^{2}$ respectively.

The overall equilibrium constants for dissociation of ${[C u {(N H_{3})}_{4}]}^{2 +} i s x * 1 0^{- 12} .$ The value of x is________ (Rounded off to the nearest integer)

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alok kumar singh

Contributor-Level 10

Stability constant are :

K₁ = 10⁴

K₂ = 1.58 * 10³

K₃ = 5 * 10²

K₄ = 10²

Overall stability constant K will be

K = K₁ * K₂ * K₃ * K₄

= 7.9 * 10¹¹

Now, overall equilibrium constant for dissociation of [Cu (NH₃)₄]²⁺ is

$K' = \frac{1}{K} = \frac{1}{7.9 * 1 0^{11}} = 0.126 * 1 0^{- 11}$

= 1.26 * 10^-12

So; x = 1 (Rounded off to the nearest integer)

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight Class 12th

Number of amphoteric compounds among the following is___________.

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alok kumar singh

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Beo & Be (OH)₂ are amphoteric in nature, because they react with both acid and base

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight Class 11th

The gas released during anaerobic degradation of vegetation may lead to:

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alok kumar singh

Contributor-Level 10

Methane gas is produced during anaerobic degradation of vegetation that leads to global warming so considered as greenhouse gas like CO₂. Also it causes cancer.

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight Biomolecules

Out of the following which type of interaction is responsible for the stabilization of a-helix structure of proteins?

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alok kumar singh

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For stabilization of a - helix structure of protein, H-bonding is responsible which is in between group present in protein.

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight Chemistry Chemical Bonding and Molecular Structure

Which of the following are isostructural pairs?

A. $S O_{4}^{2 -} a n d C r O_{4}^{2 -}$

B. SiCl₄ and TiCl₄

C. NH₃ and $N O_{3}^{-}$

D. BCl₃ and BrCl₃

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alok kumar singh

Contributor-Level 10

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight Class 12th

In the following reaction the reason why meta-nitro product also formed is:

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alok kumar singh

Contributor-Level 10

During nitration of aniline, meta product is also formed, it is because of due to presence of anilinium ion. In anilium $- \overset{+}{N} H_{3}$ group is meta directing group.

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight Chemistry Hydrocarbon

Which of the following reagent is used for the following reaction?

$C H_{3} C H_{2} C H_{3} \overset{?}{\to} C H_{3} C H_{2} C H O$

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