PRACTICAL CHEMISTRY MCQs for NEET UG CHEMISTRY — Medium, Page 2

31 Medium

Understanding +4 | -1

Which of the following statements are correct?

A. Glycerol is purified by vacuum distillation because it decomposes at its normal boiling point.

B. Aniline can be purified by steam distillation as aniline is miscible in water.

C. Ethanol can be separated from ethanol water mixture by azeotropic distillation because it forms azeotrope.

D. An organic compound is pure, if mixed M.P. is remained same.

Choose the most appropriate answer from the options given below :

View answer and explanation

Correct option: B

A, C, D only

The correct answer is B (A, C, D only) because glycerol is purified by vacuum distillation, ethanol can be separated by azeotropic distillation, a pure organic compound has the same mixed melting point, but aniline is miscible with water and cannot be purified by steam distillation.

Solution approach

Correct Answer + Core Reason: A, C, D only. Glycerol is purified by vacuum distillation, ethanol can be separated by azeotropic distillation, and a pure organic compound has the same mixed melting point. Aniline is not purified by steam distillation as it is miscible in water.

Step-by-Step Solution:

Step 1: Statement A is correct: Glycerol decomposes at its normal boiling point, so vacuum distillation is used to lower the boiling point and prevent decomposition.

Step 2: Statement B is incorrect: Aniline is miscible with water, so it cannot be purified by steam distillation. Steam distillation is used for immiscible liquids.

Step 3: Statement C is correct: Ethanol and water form an azeotrope, which can be separated by azeotropic distillation using a third component.

Step 4: Statement D is correct: A pure organic compound will have a sharp melting point, and the mixed melting point with an authentic sample will remain the same.

Why other options are wrong:
- Option A (A, B, C only): Includes incorrect statement B.
- Option C (B, C, D only): Includes incorrect statement B.
- Option D (A, B, D only): Includes incorrect statement B.

Key Concepts:
- Vacuum distillation
- Steam distillation
- Azeotropic distillation
- Melting point

Memory Aids:
- Vacuum for decomposition, Azeotrope for mixtures

Quick Tips:
- Remember the principles behind different purification techniques and their applicability.

32 Medium

32 of 50Question permalink

Remembering +4 | -1

A mixture, on heating with conc.

\mathrm{H}_2 \mathrm{SO}_4

and

\mathrm{MnO}_2

, librates brown vapour of

View answer and explanation

Correct option: A

\mathrm{Br}_2

Option A (

\mathrm{Br}_2

) is correct because the reaction of a mixture with conc. H2SO4 and MnO2 liberates brown vapors of bromine.

Solution approach

Correct Answer + Core Reason: $\mathrm{Br}_2$ . The reaction of a mixture with conc. $H_2SO_4$ and $MnO_2$ liberates brown vapors of bromine.

Step-by-Step Solution:

Step 1: Understand the Reaction: The reaction involves the oxidation of bromide ions to bromine by manganese dioxide in the presence of concentrated sulfuric acid.

Step 2: Reaction Equation:
$2NaBr + MnO_2 + 3H_2SO_4 \rightarrow MnSO_4 + 2NaHSO_4 + Br_2 + 2H_2O$

Step 3: Properties of Bromine: Bromine ( $Br_2$ ) is a reddish-brown liquid that forms brown vapors.

Step 4: Conclusion: The brown vapors are of bromine ( $Br_2$ ).

Why other options are wrong:
- Option B: $\mathrm{NO}_2$ is a reddish-brown gas, but it's produced from nitrates, not bromides, under these conditions.
- Option C: $\mathrm{HBr}$ is a colorless gas, not a brown vapor.
- Option D: $\mathrm{I}_2$ is a violet vapor, not a brown vapor.

Key Concepts:
- Oxidation of Halide Ions
- Reactions with $MnO_2$ and $H_2SO_4$

Memory Aids:
- Bromine ( $Br_2$ ) is brown.

Quick Tips:
- Remember the characteristic colors of halogens.

33 Medium

33 of 50Question permalink

Application +4 | -1

The pH range where the indicator shows change in colour are given as

(i)

4-9.7

\quad

(ii)

7.46-10.0

\quad

(iii)

6.5-4

Which of the above pH range represents titration of

I. Strong acid/strong base

\left(S_A / S_B\right)

II. Weak acid/strong base

\left(W_A / S_B\right)

III. Weak base/strong acid

\left(W_B / S_A\right)

View answer and explanation

Correct option: A

(i)

\rightarrow

I, (ii)

\rightarrow

II, (iii)

\rightarrow

III

Option A is correct because it correctly matches the pH ranges with the respective titrations based on the strength of the acid and base involved.

Solution approach

Correct Answer + Core Reason: (i) $\rightarrow$ I, (ii) $\rightarrow$ II, (iii) $\rightarrow$ III. The pH range depends on the strength of the acid and base in the titration.

Step-by-Step Solution:

Step 1: Understand the pH range for different titrations.

Step 2: Strong acid/strong base titration has a pH range around 7 (4-9.7 is acceptable due to indicator variations).

Step 3: Weak acid/strong base titration has a pH range above 7 (7.46-10.0 is appropriate).

Step 4: Weak base/strong acid titration has a pH range below 7 (6.5-4 is appropriate).

Step 5: Match the pH ranges to the corresponding titrations.

Why other options are wrong:
- Option B: Incorrectly assigns pH ranges to titrations.
- Option C: Incorrectly assigns pH ranges to titrations.
- Option D: Incorrectly assigns pH ranges to titrations.

Key Concepts:
- Acid-base titrations
- pH ranges of indicators
- Strong vs. weak acids and bases

Memory Aids:
- Strong acid/strong base: pH ~ 7
- Weak acid/strong base: pH > 7
- Weak base/strong acid: pH < 7

Quick Tips:
- Remember the general pH range for each type of titration.

34 Medium

34 of 50Question permalink

Remembering +4 | -1

In qualitative test for identification of presence of phosphorous, the compound is heated with an oxidising agent. Which is further treated with nitric acid and ammonium molybdate respectively. The yellow coloured precipitate obtained is:

View answer and explanation

Correct option: C

\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3

Option C is correct because the yellow precipitate formed in the test for phosphorus is ammonium phosphomolybdate, which has the formula (NH₄)₃PO₄ ⋅ 12MoO₃.

Solution approach

Correct Answer + Core Reason: $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3$ . The yellow precipitate formed is ammonium phosphomolybdate.

Step-by-Step Solution:

Step 1: The compound containing phosphorus is oxidized to phosphate ( $\mathrm{PO}_4^{3-}$ ).

Step 2: The phosphate ion reacts with ammonium molybdate in the presence of nitric acid to form ammonium phosphomolybdate, a yellow precipitate.

Step 3: The balanced equation for the formation of the precipitate is:
$\mathrm{H}_3\mathrm{PO}_4 + 12(\mathrm{NH}_4)_2\mathrm{MoO}_4 + 21\mathrm{HNO}_3 \rightarrow (\mathrm{NH}_4)_3\mathrm{PO}_4 \cdot 12\mathrm{MoO}_3 + 21\mathrm{NH}_4\mathrm{NO}_3 + 12\mathrm{H}_2\mathrm{O}$

Step 4: The yellow precipitate is ammonium phosphomolybdate, $(\mathrm{NH}_4)_3\mathrm{PO}_4 \cdot 12\mathrm{MoO}_3$ .

Why other options are wrong:
- Option A: $\mathrm{Na}_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3$ : This is incorrect because the precipitate formed involves ammonium molybdate, not sodium molybdate.
- Option B: $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12\left(\mathrm{NH}_4\right)_2 \mathrm{MoO}_4$ : This formula is incorrect as it includes additional ammonium molybdate molecules that are not part of the actual precipitate.
- Option D: $\mathrm{MoPO}_4 \cdot 21 \mathrm{NH}_4 \mathrm{NO}_3$ : This formula is incorrect as it does not represent the correct composition of ammonium phosphomolybdate.

Key Concepts:
- Qualitative analysis of phosphorus
- Formation of ammonium phosphomolybdate precipitate
- Chemical reactions of phosphate ions

Memory Aids:
- Remember that the yellow precipitate is ammonium phosphomolybdate: (NH₄)₃PO₄ ⋅ 12MoO₃

Quick Tips:
- In qualitative analysis questions, knowing the correct formulas of precipitates is crucial. Pay attention to the reactants and the conditions of the reaction.

35 Medium

35 of 50Question permalink

Remembering +4 | -1

\text{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]

is not used for the detection of

View answer and explanation

Correct option: A

\mathrm{Fe}^{2+}

ions

Option A (

\mathrm{Fe}^{2+}

ions) is correct because

\text{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]

reacts with

\mathrm{Fe}^{3+}

to form Prussian blue, with

\mathrm{Zn}^{2+}

to form a white precipitate, and with

\mathrm{Cu}^{2+}

to form a chocolate-brown precipitate, but it does not form a distinctive precipitate with

\mathrm{Fe}^{2+}

.

Solution approach

Correct Answer + Core Reason: $\mathrm{Fe}^{2+}$ ions. $\text{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ does not form a distinctive precipitate with $\mathrm{Fe}^{2+}$ ions.

Step-by-Step Solution:

Step 1: Consider the reactions of $\text{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ with each of the given ions.

Step 2: $\text{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ reacts with $\mathrm{Fe}^{3+}$ ions to form Prussian blue, $\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3$ .

Step 3: $\text{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ reacts with $\mathrm{Zn}^{2+}$ ions to form a white precipitate of $\mathrm{K}_2\mathrm{Zn}_3[\mathrm{Fe}(\mathrm{CN})_6]_2$ .

Step 4: $\text{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ reacts with $\mathrm{Cu}^{2+}$ ions to form a chocolate-brown precipitate of $\mathrm{Cu}_2[\mathrm{Fe}(\mathrm{CN})_6]$ .

Step 5: $\text{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ reacts with $\mathrm{Fe}^{2+}$ to form a light blue precipitate, but this is not a reliable test because it is easily oxidized to Prussian blue.

Why other options are wrong:
- Option B: $\mathrm{Zn}^{2+}$ ions form a white precipitate with $\text{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ .
- Option C: $\mathrm{Cu}^{2+}$ ions form a chocolate-brown precipitate with $\text{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ .
- Option D: $\mathrm{Fe}^{3+}$ ions form Prussian blue with $\text{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ .

Key Concepts:
- Qualitative analysis
- Complex formation

Memory Aids:
- Remember the colors of the precipitates formed with different metal ions.

Quick Tips:
- Know the characteristic reactions of common metal ions with various reagents.

36 Medium

36 of 50Question permalink

Analyzing +4 | -1

A metal salt solution forms a yellow ppt with potassium chromate in acetic acid, a white ppt with dilute sulphuric acid, but gives no ppt with sodium chloride or iodide, it is:

View answer and explanation

Correct option: C

Barium carbonate

Option C (Barium carbonate) is correct because barium ions form a yellow precipitate with potassium chromate, a white precipitate with dilute sulfuric acid, and do not precipitate with chloride or iodide ions.

Solution approach

Correct Answer + Core Reason: Barium carbonate. Barium ions give the described reactions.

Step-by-Step Solution:

Step 1: Analyze the given reactions.

Step 2: Yellow precipitate with potassium chromate in acetic acid suggests the formation of barium chromate ( ${BaCrO}_4$ ).

Step 3: White precipitate with dilute sulfuric acid suggests the formation of barium sulfate ( ${BaSO}_4$ ).

Step 4: No precipitate with sodium chloride or iodide rules out lead ( ${Pb}^{2+}$ ) and silver ( ${Ag}^{+}$ ) as they form insoluble chlorides and iodides.

Step 5: Barium ( ${Ba}^{2+}$ ) forms insoluble chromate and sulfate but soluble chloride and iodide.

Why other options are wrong:
- Option A: Lead carbonate is wrong because lead forms insoluble chlorides and iodides.
- Option B: Basic lead carbonate is wrong because lead forms insoluble chlorides and iodides.
- Option D: Strontium carbonate is wrong because strontium chromate is not as insoluble as barium chromate, and strontium sulfate is more soluble than barium sulfate.

Key Concepts:
- Solubility rules
- Precipitation reactions
- Qualitative analysis

Memory Aids:
- Remember the characteristic precipitates of barium.

Quick Tips:
- Consider the solubility rules for chromates, sulfates, chlorides, and iodides.

37 Medium

37 of 50Question permalink

Remembering +4 | -1

When a mixture of solid

{NaCl},

solid

{K}_2 {Cr}_2 {O}_7

is heated with conc.

{H}_2 {SO}_4,

orange red vapours are obtained. These are of the compound:

View answer and explanation

Correct option: B

Chromyl chloride

Option B (Chromyl chloride) is correct because the reaction of solid NaCl and solid K2Cr2O7 with conc. H2SO4 produces chromyl chloride vapors.

Solution approach

Correct Answer + Core Reason: Chromyl chloride. The reaction of solid NaCl and solid $K_2Cr_2O_7$ with conc. $H_2SO_4$ produces chromyl chloride vapors.

Step-by-Step Solution:

Step 1: Understand the Reaction: The reaction involves the formation of chromyl chloride ( $CrO_2Cl_2$ ) when a chloride salt is heated with potassium dichromate and concentrated sulfuric acid.

Step 2: Reaction Equation:
$K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 2KHSO_4 + 4NaHSO_4 + 3H_2O$

Step 3: Properties of Chromyl Chloride: Chromyl chloride is a volatile, orange-red liquid.

Step 4: Conclusion: The orange-red vapors are of chromyl chloride ( $CrO_2Cl_2$ ).

Why other options are wrong:
- Option A: Chromous chloride is not formed in this reaction.
- Option C: Chromic chloride is not formed as a vapor in this reaction.
- Option D: Chromic sulphate is not formed as a vapor in this reaction.

Key Concepts:
- Chromyl Chloride Test
- Qualitative Analysis of Chloride Ions

Memory Aids:
- Chromyl chloride ( $CrO_2Cl_2$ ) is orange-red.

Quick Tips:
- Remember the chromyl chloride test for the detection of chloride ions.

38 Medium

38 of 50Question permalink

Remembering +4 | -1

To a metal nitrate, when

\text{KI}

solution is added, a black precipitate is produced at first, on adding excess of

\text{KI},

orange solution is produced. Identify the metal ion.

View answer and explanation

Correct option: A

\mathrm{Hg}^{2+}

Option A (

\mathrm{Hg}^{2+}

) is correct because mercury(II) ions react with KI to form a black precipitate that dissolves in excess KI to form a complex.

Solution approach

Correct Answer + Core Reason: $\mathrm{Hg}^{2+}$ . Mercury(II) ions react with KI to form a black precipitate that dissolves in excess KI to form a complex.

Step-by-Step Solution:

Step 1: Reaction of Metal Nitrate with KI: The reaction involves the formation of a precipitate followed by the dissolution of the precipitate in excess KI.

Step 2: Reaction with $\mathrm{Hg}^{2+}$ : Mercury(II) ions ( $Hg^{2+}$ ) react with KI to form a red precipitate of $HgI_2$ which quickly turns black. In excess KI, the $HgI_2$ dissolves to form a colorless complex, $[HgI_4]^{2-}$ .
$Hg^{2+} + 2KI \rightarrow HgI_2 (red, turns black)\downarrow$
$HgI_2 + 2KI \rightarrow [HgI_4]^{2-} (colorless)$

Step 3: Analyze the other metal ions:
- $\mathrm{Bi}^{3+}$ : Forms a white precipitate with KI that does not dissolve in excess KI.
- $\mathrm{Cu}^{2+}$ : Forms a brown precipitate with KI.
- $\mathrm{Pb}^{2+}$ : Forms a yellow precipitate with KI that dissolves in hot water but not in excess KI.

Step 4: Conclusion: The metal ion is $\mathrm{Hg}^{2+}$ .

Why other options are wrong:
- Option B: $\mathrm{Bi}^{3+}$ forms a white precipitate that does not dissolve in excess KI.
- Option C: $\mathrm{Cu}^{2+}$ forms a brown precipitate with KI.
- Option D: $\mathrm{Pb}^{2+}$ forms a yellow precipitate that dissolves in hot water but not in excess KI.

Key Concepts:
- Qualitative Analysis
- Reactions of Metal Ions with KI
- Complex Formation

Memory Aids:
- Mercury (Hg) forms a black precipitate with KI that dissolves in excess.

Quick Tips:
- Remember the characteristic reactions of metal ions with KI.

39 Medium

39 of 50Question permalink

Remembering +4 | -1

Which of the following cations does not import colour to the flame in flame test?

View answer and explanation

Correct option: B

\mathrm{Mg}^{2+}

Option B (

\mathrm{Mg}^{2+}

) is correct because magnesium ions do not impart color to the flame in a flame test.

Solution approach

Correct Answer + Core Reason: $\mathrm{Mg}^{2+}$ . Magnesium ions do not impart color to the flame.

Step-by-Step Solution:

Step 1: Recall the principle of the flame test.

Step 2: Flame tests are used to identify certain metal ions based on the color they produce when heated in a flame.

Step 3: The color is produced when electrons in the metal ions are excited to higher energy levels and then return to their ground state, emitting light of a specific wavelength.

Step 4: $\mathrm{Mg}^{2+}$ does not have easily excitable electrons in the visible region.

Why other options are wrong:
- Option A: $\mathrm{Ba}^{2+}$ imparts a green color to the flame.
- Option C: $\mathrm{Ca}^{2+}$ imparts a brick-red color to the flame.
- Option D: $\mathrm{Cu}^{2+}$ imparts a blue-green color to the flame.

Key Concepts:
- Flame test
- Atomic emission

Memory Aids:
- Remember the characteristic colors of common metal ions in flame tests.

Quick Tips:
- Know the common flame test colors for different elements.

40 Medium

40 of 50Question permalink

Application +4 | -1

Given that the

K_a\text{(methyl orange)}=4.0 \times 10^{-4},

a solution at

\mathrm{p H}=2

containing the indicator would be

View answer and explanation

Correct option: D

Red

Option D (Red) is correct because at pH 2, which is significantly lower than the pKa of methyl orange, the indicator will be predominantly in its acidic form, which is red.

Solution approach

Correct Answer + Core Reason: Red. At pH 2, the indicator is predominantly in its acidic form, which is red.

Step-by-Step Solution:\nStep 1: Understand the relationship between pH, pKa, and indicator color.\nStep 2: Calculate the pKa of methyl orange.\nStep 3: Determine the predominant form of the indicator at pH 2.

$K_a = 4.0 \times 10^{-4}$
$pK_a = -log(K_a) = -log(4.0 \times 10^{-4}) = 3.4$ \nSince the pH (2) is much lower than the pKa (3.4), the indicator will be predominantly in its acidic form (HInd), which is red.

Why other options are wrong:
- Option A: Orange: This would be the color near the equivalence point.
- Option B: Yellow: This is the color of the basic form of the indicator.
- Option C: Colourless: Methyl orange is not colorless in either its acidic or basic form.

Key Concepts:
- Acid-base indicators
- pH and pKa

Memory Aids:
- Remember that indicators change color near their pKa.

Quick Tips:
- Use the Henderson-Hasselbalch equation to determine the ratio of the acidic and basic forms of the indicator.

41 Medium

41 of 50Question permalink

Application +4 | -1

Which of the following will be precipitated when dil.

HCl

is added to their solution.

View answer and explanation

Correct option: A

{Pb}\left({NO}_3\right)_2

Option A (

{Pb}({NO}_3)_2

) is correct because lead chloride (

PbCl_2

) is insoluble in cold dilute HCl and will precipitate out.

Solution approach

Correct Answer + Core Reason: ${Pb}({NO}_3)_2$ . Lead chloride ( $PbCl_2$ ) is insoluble in cold dilute HCl.

Step-by-Step Solution:\nStep 1: Consider the solubility rules for chlorides.\nStep 2: Determine which metal chloride is insoluble in dilute HCl.
\nLead chloride ( $PbCl_2$ ) is insoluble in cold water and dilute HCl. When dilute HCl is added to a solution of lead nitrate, lead chloride precipitates out.
$Pb(NO_3)_2(aq) + 2HCl(aq) \rightarrow PbCl_2(s) + 2HNO_3(aq)$

Why other options are wrong:
- Option B: ${CuSO}_4$ : Copper chloride is soluble in water.
- Option C: ${MnCl}_2$ : Manganese chloride is soluble in water.
- Option D: ${MgCO}_3$ : Magnesium carbonate will react with HCl to form magnesium chloride, water, and carbon dioxide, and it will dissolve.

Key Concepts:
- Solubility rules
- Precipitation reactions

Memory Aids:
- Remember that lead chloride is insoluble.

Quick Tips:
- Know the solubility rules for common ionic compounds.

42 Medium

42 of 50Question permalink

Understanding +4 | -1

Sometimes yellow turbidity appears while passing

\mathrm{H}_2 \mathrm{S}

gas even in the absence of II group radicals. This is because of

View answer and explanation

Correct option: C

The oxidation of

\mathrm{H}_2 \mathrm{S}

gas by some acid radicals

Option C (The oxidation of

\mathrm{H}_2 \mathrm{S}

gas by some acid radicals) is correct because oxidizing agents present in the solution can oxidize hydrogen sulfide to elemental sulfur, causing a yellow turbidity.

Solution approach

Correct Answer + Core Reason: The oxidation of $\mathrm{H}_2 \mathrm{S}$ gas by some acid radicals. Oxidizing agents oxidize $H_2S$ to elemental sulfur, causing turbidity.

Step-by-Step Solution:
Step 1: $H_2S$ gas is passed through a solution containing metal ions for qualitative analysis.
Step 2: If oxidizing agents are present (e.g., $Fe^{3+}$ , $NO_3^−$ ), they can oxidize $H_2S$ to elemental sulfur (S).
Step 3: The reaction is: $2H_2S + O_2 \rightarrow 2S + 2H_2O$ (where $O_2$ represents an oxidizing agent).
Step 4: The elemental sulfur is insoluble and forms a colloidal suspension, causing a milky or yellow turbidity.

Why other options are wrong:
- Option A: Sulphur is present in the mixture as impurity: While sulfur could be an impurity, the turbidity is more likely due to the oxidation of $H_2S$ .
- Option B: IV group radicals are precipitated as sulphides: If IV group radicals were present, they would precipitate as colored sulfides, not as a yellow turbidity.
- Option D: III group radicals are precipitated as hydroxides: III group radicals precipitate as hydroxides under basic conditions, not during the passage of $H_2S$ in acidic conditions.

Key Concepts:
- Oxidation-reduction reactions
- Solubility of sulfur
- Qualitative analysis

Memory Aids:
- Remember that oxidizing agents can oxidize $H_2S$ to sulfur.

Quick Tips:
- Consider the presence of oxidizing agents when performing sulfide precipitations.

43 Medium

43 of 50Question permalink

Analyzing +4 | -1

A substance on treatment with dil.

\mathrm{H}_2 \mathrm{SO}_4

liberates a colourless gas which produces (i) turbidity with baryta water and (ii) turns acidified dichromate solution green. The reaction indicates the presence of

View answer and explanation

Correct option: C

\mathrm{SO}_3^{2-}

Option C is correct because sulfite ions (SO3^2-) react with dilute H2SO4 to release sulfur dioxide (SO2), which turns baryta water turbid and acidified dichromate solution green.

Solution approach

Correct Answer + Core Reason: $\mathrm{SO}_3^{2-}$ . The gas liberated is $\mathrm{SO}_2$ , which produces turbidity with baryta water and turns acidified dichromate green.

Step-by-Step Solution:

Step 1: Consider the reaction of each anion with dilute $\mathrm{H}_2\mathrm{SO}_4$ .

Step 2: $\mathrm{CO}_3^{2-} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{SO}_4^{2-}$ . $\mathrm{CO}_2$ does not turn acidified dichromate green.

Step 3: $\mathrm{S}^{2-} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{H}_2\mathrm{S} + \mathrm{SO}_4^{2-}$ . $\mathrm{H}_2\mathrm{S}$ turns lead acetate paper black, not dichromate green.

Step 4: $\mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{SO}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{SO}_4^{2-}$ . $\mathrm{SO}_2$ turns baryta water turbid and acidified dichromate green.

Step 5: $\mathrm{NO}_2^{-} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{HNO}_2$ . This reaction is more complex and doesn't directly produce a gas with the given properties.

Step 6: Therefore, the anion is $\mathrm{SO}_3^{2-}$ .

Why other options are wrong:
- Option A: $\mathrm{CO}_3^{2-}$ produces $\mathrm{CO}_2$ , which doesn't turn dichromate green.
- Option B: $\mathrm{S}^{2-}$ produces $\mathrm{H}_2\mathrm{S}$ , which turns lead acetate paper black.
- Option D: $\mathrm{NO}_2^{-}$ does not directly produce a gas with the given properties.

Key Concepts:
- Qualitative analysis
- Reactions of anions with acids
- Properties of gases like $\mathrm{SO}_2$ and $\mathrm{CO}_2$

Memory Aids:
- Remember the characteristic reactions of common anions.

Quick Tips:
- Consider the gases produced by each anion and their properties.

44 Medium

44 of 50Question permalink

Analyzing +4 | -1

Given below are two statements one is labelled as Assertion (A) and the other is labeilled as Reason.

Assertion (A): In the titration of HCl against NaOH, phenolphthalein is used as suitable indicator.

Reason (R): Phenolphthalein is pink coloured in basic medium.

In the light of above statements, choose the most appropriate answer from the options given below.

View answer and explanation

Correct option: D

If both (A) and (R) are correct, but (R) is not the
correct explanation of (A)

Option D is correct because both the assertion and reason are correct, but the reason does not explain why phenolphthalein is a suitable indicator for the titration of HCl against NaOH.

Solution approach

Correct Answer + Core Reason: If both (A) and (R) are correct, but (R) is not the correct explanation of (A). Phenolphthalein is suitable for strong acid-strong base titrations, and it is pink in basic medium, but the latter doesn't explain the former.

Step-by-Step Solution:

Step 1: Analyze the assertion: In the titration of HCl against NaOH, phenolphthalein is a suitable indicator. This is true because the pH at the equivalence point is around 7, which falls within the pH range of phenolphthalein.

Step 2: Analyze the reason: Phenolphthalein is pink colored in basic medium. This is also true.

Step 3: Determine if the reason explains the assertion. The fact that phenolphthalein is pink in basic medium doesn't explain why it's suitable for this specific titration. Its suitability depends on its pH range aligning with the equivalence point of the titration.

Why other options are wrong:
- Option A: The reason is not the correct explanation.
- Option B: Both assertion and reason are correct.
- Option C: Both assertion and reason are correct.

Key Concepts:
- Acid-base titrations
- Acid-base indicators
- Equivalence point
- pH range of indicators

Memory Aids:
- Remember that the choice of indicator depends on the pH at the equivalence point.

Quick Tips:
- Consider the pH range of the indicator and the pH at the equivalence point of the titration.

45 Medium

45 of 50Question permalink

Understanding +4 | -1

Which of the following nitrogen containing compound does not give Lassaigne's test?

View answer and explanation

Correct option: D

Hydrazine

The correct answer is D (Hydrazine) because hydrazine (

N_2H_4

) does not contain carbon, which is required for the Lassaigne's test to form cyanide ions.

Solution approach

Correct Answer + Core Reason: Hydrazine. Hydrazine does not give Lassaigne's test because it lacks carbon.

Step-by-Step Solution:

Step 1: Lassaigne's test requires the organic compound to contain carbon, nitrogen, sulfur, or halogens.

Step 2: The organic compound is fused with sodium metal to convert these elements into ionic forms.

Step 3: For nitrogen to be detected, it must be converted to sodium cyanide ( $NaCN$ ). This requires the presence of carbon in the original compound.

Step 4: Hydrazine ( $N_2H_4$ ) contains only nitrogen and hydrogen, lacking carbon. Therefore, it cannot form cyanide ions and will not give a positive Lassaigne's test for nitrogen.

Why other options are wrong:
- Option A (Phenyl hydrazine): Phenyl hydrazine contains carbon, nitrogen, and hydrogen and will give a positive Lassaigne's test.
- Option B (Glycene): Glycene (Glycine) contains carbon, nitrogen, oxygen, and hydrogen and will give a positive Lassaigne's test.
- Option C (Urea): Urea contains carbon, nitrogen, oxygen, and hydrogen and will give a positive Lassaigne's test.

Key Concepts:
- Lassaigne's test
- Carbon requirement
- Hydrazine structure

Memory Aids:
- No Carbon, No Lassaigne's test (for nitrogen)

Quick Tips:
- Remember that carbon is essential for a positive Lassaigne's test for nitrogen.

46 Medium

46 of 50Question permalink

Understanding +4 | -1

Given below are two statements.

Statement (I) : The gas liberated on warming a salt with dil

\mathrm{H}_2 \mathrm{SO}_4,

turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion.

Statement (II) : In statement-I the colour of paper turns black because of formation of lead sulphite.

In the light of the above statements, choose the most appropriate answer from the options given below

View answer and explanation

Correct option: B

Statement I is true but Statement II is false

Option B (Statement I is true but Statement II is false) is correct because hydrogen sulfide gas turns lead acetate paper black due to the formation of lead sulfide, not lead sulfite.

Solution approach

Correct Answer + Core Reason: Statement I is true but Statement II is false. $H_2S$ reacts with lead acetate to form lead sulfide, which is black.

Step-by-Step Solution:
Step 1: Statement I describes the reaction of a sulfide salt with dilute $H_2SO_4$ to produce $H_2S$ gas.
Step 2: $H_2S$ gas reacts with lead acetate ( $Pb(CH_3COO)_2$ ) on the paper to form lead sulfide ( $PbS$ ).
Step 3: The reaction is: $Pb(CH_3COO)_2(aq) + H_2S(g) \rightarrow PbS(s) + 2CH_3COOH(aq)$
Step 4: Lead sulfide ( $PbS$ ) is a black precipitate.
Step 5: Statement II claims the color is due to lead sulfite, which is incorrect. The color is due to lead sulfide.

Why other options are wrong:
- Option A: Both Statement I and Statement II are false: Statement I is true, so this option is incorrect.
- Option C: Both Statement I and Statement II are true: Statement II is false, so this option is incorrect.
- Option D: Statement I is false but Statement II is true: Statement I is true, so this option is incorrect.

Key Concepts:
- Qualitative analysis of sulfide ions
- Reaction of sulfides with lead acetate
- Formation of precipitates

Memory Aids:
- Remember that $H_2S$ turns lead acetate paper black due to the formation of $PbS$ .

Quick Tips:
- Know the characteristic reactions and colors of common ions in qualitative analysis.

47 Medium

47 of 50Question permalink

Understanding +4 | -1

While preparing crystals of Mohr's salt, dil.

\mathrm{H}_2 \mathrm{SO}_4

is added to a mixture of ferrous sulphate and ammonium sulphate, before dissolving this mixture in water, dil.

\mathrm{H}_2 \mathrm{SO}_4

is added here to:

View answer and explanation

Correct option: A

Prevent the hydrolysis of ferrous sulphate

Option A (Prevent the hydrolysis of ferrous sulphate) is correct because dilute sulfuric acid is added to prevent the hydrolysis of ferrous sulfate by maintaining a low pH.

Solution approach

Correct Answer + Core Reason: Prevent the hydrolysis of ferrous sulphate. Dilute sulfuric acid prevents the hydrolysis of $Fe^{2+}$ ions by maintaining a low pH.

Step-by-Step Solution:
Step 1: Ferrous sulfate ( $FeSO_4$ ) can undergo hydrolysis in water, forming iron hydroxide and sulfuric acid: $Fe^{2+}(aq) + 2H_2O(l) \rightleftharpoons Fe(OH)_2(s) + 2H^+(aq)$
Step 2: The formation of iron hydroxide can lead to a precipitate and reduce the yield of Mohr's salt.
Step 3: Adding dilute sulfuric acid ( $H_2SO_4$ ) increases the concentration of $H^+$ ions in the solution.
Step 4: This shifts the equilibrium to the left, suppressing the hydrolysis of $Fe^{2+}$ ions and preventing the formation of iron hydroxide.

Why other options are wrong:
- Option B: Prevent the hydrolysis of ammonium sulphate: Ammonium sulfate is less prone to hydrolysis than ferrous sulfate, so this is not the primary reason.
- Option C: Make the medium strongly acidic: While the medium becomes acidic, the main purpose is to prevent hydrolysis, not just to make it strongly acidic.
- Option D: Increase the rate of formation of crystals: The addition of dilute sulfuric acid primarily prevents hydrolysis, not directly increasing the rate of crystal formation.

Key Concepts:
- Hydrolysis of metal ions
- Le Chatelier's principle
- Common ion effect
- Preparation of Mohr's salt

Memory Aids:
- Remember that acids suppress the hydrolysis of metal ions.

Quick Tips:
- Understand the factors that affect the hydrolysis of metal ions.

48 Medium

48 of 50Question permalink

Remembering +4 | -1

Disodium hydrogen phoshate is used to test:

View answer and explanation

Correct option: A

\mathrm{Mg}^{2+}

Option A (

\mathrm{Mg}^{2+}

) is correct because disodium hydrogen phosphate is used as a reagent to test for magnesium ions.

Solution approach

Correct Answer + Core Reason: $\mathrm{Mg}^{2+}$ . Disodium hydrogen phosphate is used to test for $\mathrm{Mg}^{2+}$ ions.

Step-by-Step Solution:

Step 1: Recall the reaction of $\mathrm{Mg}^{2+}$ with disodium hydrogen phosphate.

Step 2: $\mathrm{Mg}^{2+}$ ions react with disodium hydrogen phosphate ( $\mathrm{Na_2HPO_4}$ ) in the presence of ammonia to form a white precipitate of magnesium ammonium phosphate ( $\mathrm{MgNH_4PO_4}$ ).

Step 3: The formation of this precipitate indicates the presence of $\mathrm{Mg}^{2+}$ ions.

Why other options are wrong:
- Option B: $\mathrm{Na}^{+}$ does not form a precipitate with disodium hydrogen phosphate.
- Option C: $\mathrm{Ca}^{2+}$ may form a precipitate with phosphate, but the test is more specific for $\mathrm{Mg}^{2+}$ in the presence of ammonia.
- Option D: All - This is incorrect because the test is primarily used for $\mathrm{Mg}^{2+}$ .

Key Concepts:
- Qualitative analysis
- Precipitation reactions

Memory Aids:
- $\mathrm{Mg}^{2+}$ + $\mathrm{Na_2HPO_4}$ + $\mathrm{NH_3}$ → $\mathrm{MgNH_4PO_4}$ (white precipitate).

Quick Tips:
- Remember the specific reagents used to test for different ions.

49 Medium

49 of 50Question permalink

Understanding +4 | -1

In copper turnings test, pieces of copper is added to a nitrate salt and heated with conc.

\mathrm{H}_2 \mathrm{SO}_4.

In this reaction copper acts as

View answer and explanation

Correct option: C

Reducing agent

Option C (Reducing agent) is correct because copper loses electrons (oxidation) to reduce nitrate ions to nitrogen dioxide.

Solution approach

Correct Answer + Core Reason: Reducing agent. Copper loses electrons and reduces nitrate ions.

Step-by-Step Solution:

Step 1: The reaction involves copper turnings, a nitrate salt, and concentrated sulfuric acid.

Step 2: The nitrate salt is reduced to nitrogen dioxide ( $NO_2$ ).

Step 3: Copper is oxidized to copper(II) ions ( $Cu^{2+}$ ).

Step 4: The half-reactions are:
- Oxidation: $Cu \rightarrow Cu^{2+} + 2e^-$
- Reduction: $NO_3^- + 2H^+ + e^- \rightarrow NO_2 + H_2O$

Step 5: Since copper loses electrons, it acts as a reducing agent.

Why other options are wrong:
- Option A: Oxidising agent: Copper is oxidized, not reduced, so it's not an oxidizing agent.
- Option B: Dehydrating agent: Concentrated sulfuric acid acts as a dehydrating agent, not copper.
- Option D: Bleaching agent: Copper does not act as a bleaching agent in this reaction.

Key Concepts:
- Redox reactions
- Oxidizing and reducing agents

Memory Aids:
- Oxidation is loss of electrons (LEO), reduction is gain of electrons (GER).

Quick Tips:
- Identify the oxidation states of the reactants and products to determine the oxidizing and reducing agents.

50 Medium

50 of 50Question permalink

Application +4 | -1

At what pH will a

10^{-3} \mathrm{M}

solution of indicator with

K_b=10^{-10}

changes colour

View answer and explanation

Correct option: B

4

Option B (4) is correct because the pH at which an indicator changes color is approximately equal to its pK

_a

value. Since K

_b

is given, we can calculate pK

_a

and thus the pH.

Solution approach

Correct Answer + Core Reason: 4. The pH at which the indicator changes color is approximately equal to its pK $_a$ .

Step-by-Step Solution:

Step 1: Understand the problem. We are given the K $_b$ of an indicator and its concentration, and we need to find the pH at which it changes color.

Step 2: Recall the relationship between K $_a$ and K $_b$ . We know that $K_a \times K_b = K_w$ , where $K_w = 10^{-14}$ .

Step 3: Calculate K $_a$ . $K_a = \frac{{K_w}}{{K_b}} = \frac{{10^{-14}}}{{10^{-10}}} = 10^{-4}$

Step 4: Calculate pK $_a$ . $pK_a = -\log(K_a) = -\log(10^{-4}) = 4$

Step 5: Determine the pH. The pH at which the indicator changes color is approximately equal to its pK $_a$ . Therefore, pH = 4.

Final Answer: 4

Why other options are wrong:
- Option A (10): This is the pK $_b$ value, not the pK $_a$ value. The indicator changes color around its pK $_a$ , not pK $_b$ .
- Option C (3): This value is incorrect. It might be a result of miscalculating the pK $_a$ or misunderstanding the relationship between K $_a$ , K $_b$ , and pH.
- Option D (7): This is the pH of a neutral solution. Indicators change color at different pH values depending on their K $_a$ and K $_b$ values.

Key Concepts:
- Acid-base indicators
- K $_a$ , K $_b$ , and K $_w$ relationship: $K_a \times K_b = K_w$
- pK $_a$ and pH relationship

Memory Aids:
- $pK_a = -\log(K_a)$
- $pH = pK_a$ at the equivalence point

Quick Tips:
- Remember the relationship between K $_a$ , K $_b$ , and K $_w$ .
- The indicator changes color at a pH close to its pK $_a$ .

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