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In the following structure, the double bonds are marked as I, II, III and IV Geometrical isomerism is not possible at site (s) :
Option: 1 III
Option: 2 I
Option: 3 I and II
Option: 4 III and IV

Geometrical isomerism is not possible at Site I as two identical methyl groups are attached to the same carbon bearing the double bond.

Hence, the answer is Option (2)

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Posted by

vishal kumar

What quantity (in mL) of a 45% acid solution of a monoprotic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution?
Option: 1 320
Option: 2 325
Option: 3 316
Option: 4 330

Reactions in Solutions -

Concentration :
It is the amount of solute present in one litre of solution. It is denoted by C or S.

$\\ \mathrm{C\, or\, S\, =\, \frac{Weight\, of\, solute\, in\, gram}{Volume\, in\, litre}} \\\\ \mathrm{C\, =\, N\, \times\, E}\\\\\mathrm{Here\, N\, =\, normality\, and \, E\, =\, Eq.\, wt.}$

Mole Fraction:
It is the ratio of moles of one component to the total number of moles present In the solution. It is expressed by X for example, for a binary solution with two components A and B.

$X_{A}\, = \frac{n_{A}}{n_{A}\, +\, n_{B}}$

$X_{B}\, = \frac{n_{B}}{n_{A}\, +\, n_{B}}$

$X_{A}\, +\, X_{B}\, =\, 1$

Here n_A and n_B represent moles of solvent and solute respectively. Mole fraction does not depend upon temperature as both solute and solvent are expressed by weight.

Molarity:
It is the number of moles or gram moles of solute dissolved per litre of the solution. Molarity is denoted by 'M'.

$\mathrm{M\, =\, \frac{Weight\, of\, solute\, in\, gram}{Molar\, mass\, \times\, volume\, in\, litre}}$

When molarity of a solution is one, it is called a molar solution and when it is 0.1, solution is called decimolar solution.
Molarity depends upon temperature and its unit is mol/litre.
M₁V₁ = M₂V₂
On dilution water added = V₂ - V₁
When the solution of two different substances react together then
$\mathrm{\frac{M_{1}V_{1}}{n_{1}}\, =\, \frac{M_{2}V_{2}}{n_{2}}}$
Here M, V, n are molarity, volume and number of molecules taking part in a reaction respectively.
When a mixture of different solutions having different concentrations are taken the molarity of the mixture is calculated as follows:
$\mathrm{M\, =\, \frac{M_{1}V_{1}\, + M_{2}V_{2}\,.....}{V_{1}\, +\, V_{2}.....}}$
When density and % by weight of a substance in a solution are given, molarity is find as follows:
$\mathrm{M\, =\,\frac{\%\, by\, weight\,\times\, d\, \times\, 10}{Molecular\, weight}}$
Here d = density

Molality
It is the number of moles or gram moles of solute dissolved per kilogram of the solvent. It is denoted by 'm'.
$\mathrm{m\, =\, \frac{Weight\, of\, solute\, in\, gram }{Molar\, mass\, \times\, wt.\, of\, solvent\, in\, Kg}}$

If molality is one solution, it is called molal solution.
One molal solution is less than one molar solution.
Molality is preferred over molarity during experiments as molality is temperature independent while molarity is temperature-dependent.

Normality
It is the number of gram equivalents of solute present in one litre of the solution and it is denoted by 'N'.
$\mathrm{N\, =\, \frac{Weight\, of\, solute\, in\, gram }{Equivalent\, mass\, \times\, volume\, in\, litre }}$

When normality of a solution is one, the solution is called normal solution and when it is 0.1, the solution is called deci-normal solution.

Normality Equation:

$\mathrm{N_{1}V_{1}\, =\, N_{2}V_{2}}$

Volume Of water added = V₂ - V₁
Here V₂ = volume after dilution
V₁ = volume before dilution
When density and % by weight of a substance in a solution are given, normality is find as follows:

$\mathrm{N\, =\, \frac{\%\, by\, weight\, \times\, d \times\, 10 }{Equivalent\ weight }}$
Here d = density of solution
When a mixture of different solutions having different concentrations are taken the normality of the mixture is calculated as follows:
$\mathrm{N\, =\, \frac{N_{1}V_{1}\, + N_{2}V_{2}\,.....}{V_{1}\, +\, V_{2}.....}}$
• In case of acid-base neutralization the normality of the resulting solution
$\mathrm{N\, =\, \frac{N_{1}V_{1}\, - \, N_{2}V_{2}\,.....}{V_{1}\, +\, V_{2}.....}}$
• To find weight of substance
$W\, =\, \frac{NEV}{1000}$

Relation between Normality and Molarity :
N x Eq wt. = molarity x molar mass
N = molarity x valency
N = molarity x number of H⁺ or OH^- ion

-

As we have learnt,

$\\\mathrm{\frac{V\times45}{100}\,+\,\frac{(800-V)20}{100}\, =\, \frac{800\times29.875}{100}}\\\\\mathrm{\frac{9V}{20}\, +\, 160\, -\, \frac{V}{5}\, =\, 239}\\\\\mathrm{\frac{5V}{20}\, =\, 79\,}\\\\\mathrm{\therefore V\, =\, 316}$

Therefore, Option (3) is correct

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Posted by

vishal kumar

Excess of NaOH(aq) was added to 100 mL of FeCl₃(aq) resulting into 2.14 g of Fe(OH)₃. The molarity of FeCl₃(aq) is :
(Given molar mass of Fe=56 g mol⁻¹ and molar mass of Cl=35.5 g mol⁻¹)
Option: 1 0.2 M
Option: 2 0.3 M
Option: 3 0.6 M
Option: 4 1.8 M

Reactions in Solutions -

Concentration :
It is the amount of solute present in one litre of solution. It is denoted by C or S.

$\\ \mathrm{C\, or\, S\, =\, \frac{Weight\, of\, solute\, in\, gram}{Volume\, in\, litre}} \\\\ \mathrm{C\, =\, N\, \times\, E}\\\\\mathrm{Here\, N\, =\, normality\, and \, E\, =\, Eq.\, wt.}$

Mole Fraction:
It is the ratio of moles of one component to the total number of moles present In the solution. It is expressed by X for example, for a binary solution with two components A and B.

$X_{A}\, = \frac{n_{A}}{n_{A}\, +\, n_{B}}$

$X_{B}\, = \frac{n_{B}}{n_{A}\, +\, n_{B}}$

$X_{A}\, +\, X_{B}\, =\, 1$

Here n_A and n_B represent moles of solvent and solute respectively. Mole fraction does not depend upon temperature as both solute and solvent are expressed by weight.

Molarity:
It is the number of moles or gram moles of solute dissolved per litre of the solution. Molarity is denoted by 'M'.

$\mathrm{M\, =\, \frac{Weight\, of\, solute\, in\, gram}{Molar\, mass\, \times\, volume\, in\, litre}}$

When molarity of a solution is one, it is called a molar solution and when it is 0.1, solution is called decimolar solution.
Molarity depends upon temperature and its unit is mol/litre.
M₁V₁ = M₂V₂
On dilution water added = V₂ - V₁
When the solution of two different substances react together then
$\mathrm{\frac{M_{1}V_{1}}{n_{1}}\, =\, \frac{M_{2}V_{2}}{n_{2}}}$
Here M, V, n are molarity, volume and number of molecules taking part in a reaction respectively.
When a mixture of different solutions having different concentrations are taken the molarity of the mixture is calculated as follows:
$\mathrm{M\, =\, \frac{M_{1}V_{1}\, + M_{2}V_{2}\,.....}{V_{1}\, +\, V_{2}.....}}$
When density and % by weight of a substance in a solution are given, molarity is find as follows:
$\mathrm{M\, =\,\frac{\%\, by\, weight\,\times\, d\, \times\, 10}{Molecular\, weight}}$
Here d = density

Molality
It is the number of moles or gram moles of solute dissolved per kilogram of the solvent. It is denoted by 'm'.
$\mathrm{m\, =\, \frac{Weight\, of\, solute\, in\, gram }{Molar\, mass\, \times\, wt.\, of\, solvent\, in\, Kg}}$

If molality is one solution, it is called molal solution.
One molal solution is less than one molar solution.
Molality is preferred over molarity during experiments as molality is temperature independent while molarity is temperature-dependent.

Normality
It is the number of gram equivalents of solute present in one litre of the solution and it is denoted by 'N'.
$\mathrm{N\, =\, \frac{Weight\, of\, solute\, in\, gram }{Equivalent\, mass\, \times\, volume\, in\, litre }}$

When normality of a solution is one, the solution is called normal solution and when it is 0.1, the solution is called deci-normal solution.

Normality Equation:

$\mathrm{N_{1}V_{1}\, =\, N_{2}V_{2}}$

Volume Of water added = V₂ - V₁
Here V₂ = volume after dilution
V₁ = volume before dilution
When density and % by weight of a substance in a solution are given, normality is find as follows:

$\mathrm{N\, =\, \frac{\%\, by\, weight\, \times\, d \times\, 10 }{Equivalent\ weight }}$
Here d = density of solution
When a mixture of different solutions having different concentrations are taken the normality of the mixture is calculated as follows:
$\mathrm{N\, =\, \frac{N_{1}V_{1}\, + N_{2}V_{2}\,.....}{V_{1}\, +\, V_{2}.....}}$
• In case of acid-base neutralization the normality of the resulting solution
$\mathrm{N\, =\, \frac{N_{1}V_{1}\, - \, N_{2}V_{2}\,.....}{V_{1}\, +\, V_{2}.....}}$
• To find weight of substance
$W\, =\, \frac{NEV}{1000}$

Relation between Normality and Molarity :
N x Eq wt. = molarity x molar mass
N = molarity x valency
N = molarity x number of H⁺ or OH^- ion

-

The chemical equation for reaction is as follows:

3NaOH(aq) + FeCl₃(aq) —> 3NaCl(aq) + F(OH)₃(aq)

Therefore, the mole ratio of FeCl₃ and Fe(OH)₃ is 1 : 1

Now moles of Fe(OH)₃ is given by:
Moles = Given mass / Molar mass

Molar mass of Fe(OH)₃= 56 + 48 +3 = 107 grams / moles

2.14 g / (107g / mole) = 0.02 moles.

Therefore, the moles for FeCl₃ is also 0.02 moles

Now, Molarity = moles per volume(L)

Molarity of FeCl₃ is thus : (0.02 / 100) x 1000

Therefore, molarity of FeCl₃ = 0.2M

Thus, option (1) is correct

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Posted by

vishal kumar

The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight (in kg) which a 75 kg person would gain if all $^{1}H$ atoms are replaced by $^{2}H$ atoms is :
Option: 1 7.5
Option: 2 10
Option: 3 15
Option: 4 37.5

Given that

Mass of the person = 75 kg

Mass of ₁H¹ present in person = 10% of 75 kg = 7.5 kg

Since Mass of ₁H² is double the Mass of ₁H¹

So, Mass of ₁H² will be in person = 2 X 7.5 kg =15 kg

Thus, increase in weight = 15 - 7.5 = 7.5 kg

Therefore, Option (1) is correct

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Posted by

vishal kumar

Which of the following compounds will show the maximum 'enol' content?
Option: 1 $CH_{3}COCH_{2}COOC_{2}H_{5}$
Option: 2 $CH_{3}COCH_{2}COCH_{3}$
Option: 3 $CH_{3}COCH_{3}$
Option: 4 $CH_{3}COCH_{2}CONH_{2}$

Difinition of Tautomerism -

Tautomers are isomers of a compound that differ only in the position of the protons and electrons. The carbon skeleton of the compound is unchanged. A reaction that involves simple proton transfer in an intramolecular fashion is called tautomerism.

The stability of enol depends on the factors

(1) Resonance

(2) Hydrogen-bonding

(3) Hyperconjugation

(4) Hydrogen which is removed from $\alpha$ - carbon should be acidic in Nature for enol formation.

CH₂ is present between 2- electron-withdrawing group, It is having acid 'H'

Over the resonance effect decreases because of cross conjugation. So the stability of the enol form decreases.

The two -dicarbonyl compounds have a higher enol content than the two mono carbonyl compounds because hydrogen bonding and conjugation stabilize their enols. The enol content in C (a mono aldehyde) is higher than D because of the reasons outlined above.

Therefore, option (2) is correct.

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Posted by

Ritika Jonwal

Among the compounds, A and B with molecular formula $C_{9}H_{18}O_{3},$ A is having higher boiling point than B. The possible structures of A and B are :
Option: 1
Option: 2
Option: 3
Option: 4

In (A), extensive inter-molecular H-bonding is possible while in (B) there is no Inter-molecular H-bonding.

Option 4 has this type of arrangement that follows the above conditions.

Therefore, Option(4) is correct.

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Posted by

Kuldeep Maurya

Which of the following compounds show geometrical isomerisation?

Option: 1 4-Methylpent-1-ene
Option: 2 2-Methylpent-1-ene
Option: 3 2-Methylpent-2-ene
Option: 4 4-Methylpent-2-ene

4-Methylpent-2-ene shows geometrical isomerism.

Therefore, Option(4) is correct.

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Posted by

Kuldeep Maurya

The increasing order of $\mathrm{pK_{b }}$ values of the following compounds is :
Option: 1 II<IV<III<I
Option: 2 II<I<III<IV
Option: 3I<II<III<IV
Option: 4 I<II<IV<III

III contains -M and -I group, hence it is the least basic.

IV contains -I group

I contains +M group and is more basic than II.

Thus order of strength: I > II > IV > III.

Order of $\mathrm{pK_{b}}$ : I < II < IV < III.

Therefore, Option(4) is correct.

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Posted by

Kuldeep Maurya

The correct order of statbility for the following alkoxides is:
Option: 1 C > A > B
Option: 2 C > B > A
Option: 3 B > A > C
Option: 4 B > C > A

When a negative charge is delocalised with an electron-withdrawing group like (NO₂) then stability increases.

(A) The negative charge is localised

(B) The negative charge is delocalised with the carbon of the alkene

(C) Negative charge is delocalised with NO₂ group

So, the order will be

C > B > A

Therefore, Option(2) is correct.

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Posted by

Ritika Jonwal

The ammonia $(NH_{3})$ released on quantitative reaction of 0.6g, urea $(NH_{2}CONH_{2})$ with sodium hydroxide $(NaOH)$ can be neutralised by:
Option: 1 200 ml of 0.2 N HCl
Option: 2200 ml of 0.4 N HCl
Option: 3100 ml of 0.1N HCl
Option: 4100 ml of 0.2N HCl

2 × mole of Urea = mole of $NH_{3}$ ........(1)
mole of $NH_{3}$ = mole of $HCl$ ........(2)
$\therefore$ mole of HCl = 2 × mole of Urea

mole of HCl = $2\times \frac{0.6}{60}=0.02mol$ ...(i)

[We know , mole = M X V = N X n X V]

$\mathrm{100 \ ml\times 0.2N\times 1=0.02\ mol}$ ...as (i).

Therefore, Option(4) is correct.

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In the following structure, the double bonds are marked as I, II, III and IV Geometrical isomerism is not possible at site (s) : Option: 1 III Option: 2 I Option: 3 I and II Option: 4 III and IV

Posted by

vishal kumar

What quantity (in mL) of a 45% acid solution of a monoprotic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution? Option: 1 320 Option: 2 325 Option: 3 316 Option: 4 330

Posted by

vishal kumar

Crack CUET with india's "Best Teachers"

Excess of NaOH(aq) was added to 100 mL of FeCl3(aq) resulting into 2.14 g of Fe(OH)3. The molarity of FeCl3(aq) is : (Given molar mass of Fe=56 g mol−1 and molar mass of Cl=35.5 g mol−1) Option: 1 0.2 M Option: 2 0.3 M Option: 3 0.6 M Option: 4 1.8 M

Posted by

vishal kumar

The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight (in kg) which a 75 kg person would gain if all atoms are replaced by atoms is : Option: 1 7.5 Option: 2 10 Option: 3 15 Option: 4 37.5

Posted by

vishal kumar

JEE Main high-scoring chapters and topics

Which of the following compounds will show the maximum 'enol' content? Option: 1 Option: 2 Option: 3 Option: 4

In the following structure, the double bonds are marked as I, II, III and IV Geometrical isomerism is not possible at site (s) :
Option: 1 III
Option: 2 I
Option: 3 I and II
Option: 4 III and IV

What quantity (in mL) of a 45% acid solution of a monoprotic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution?
Option: 1 320
Option: 2 325
Option: 3 316
Option: 4 330

Excess of NaOH(aq) was added to 100 mL of FeCl₃(aq) resulting into 2.14 g of Fe(OH)₃. The molarity of FeCl₃(aq) is :
(Given molar mass of Fe=56 g mol⁻¹ and molar mass of Cl=35.5 g mol⁻¹)
Option: 1 0.2 M
Option: 2 0.3 M
Option: 3 0.6 M
Option: 4 1.8 M

Which of the following compounds will show the maximum 'enol' content?
Option: 1 $CH_{3}COCH_{2}COOC_{2}H_{5}$
Option: 2 $CH_{3}COCH_{2}COCH_{3}$
Option: 3 $CH_{3}COCH_{3}$
Option: 4 $CH_{3}COCH_{2}CONH_{2}$

Among the compounds, A and B with molecular formula $C_{9}H_{18}O_{3},$ A is having higher boiling point than B. The possible structures of A and B are :
Option: 1
Option: 2
Option: 3
Option: 4

Which of the following compounds show geometrical isomerisation?

Option: 1 4-Methylpent-1-ene
Option: 2 2-Methylpent-1-ene
Option: 3 2-Methylpent-2-ene
Option: 4 4-Methylpent-2-ene

The increasing order of $\mathrm{pK_{b }}$ values of the following compounds is :
Option: 1 II<IV<III<I
Option: 2 II<I<III<IV
Option: 3I<II<III<IV
Option: 4 I<II<IV<III

The correct order of statbility for the following alkoxides is:
Option: 1 C > A > B
Option: 2 C > B > A
Option: 3 B > A > C
Option: 4 B > C > A

The ammonia $(NH_{3})$ released on quantitative reaction of 0.6g, urea $(NH_{2}CONH_{2})$ with sodium hydroxide $(NaOH)$ can be neutralised by:
Option: 1 200 ml of 0.2 N HCl
Option: 2200 ml of 0.4 N HCl
Option: 3100 ml of 0.1N HCl
Option: 4100 ml of 0.2N HCl