JEE Chemistry Salt Analysis MCQs Set 05

Chemistry Objective Questions and Answers: Salt Analysis

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Multiple Choice Questions

Question 1: An aqueous solution of a salt X turns blood red on treatment with CNS− and blue on treatment with K4[Fe(CN)6]. X also gives a positive chromyl chloride test. The salt X is:
(a) CuCl2
(b) FeCl3
(c) Cu(NO3)2
(d) Fe(NO3)3
Show Answer & Explanation

Answer: (b) FeCl3

Explanation:
1. Blood red colour with CNS− and Prussian blue with K4[Fe(CN)6] both indicate Fe3+.
2. A positive chromyl chloride test indicates Cl−, so the salt is FeCl3.
3. Answer: (b)

Teacher's Note:
1. Fe3+ gives Fe(SCN) red complex and Prussian blue.
2. Chromyl chloride test is specific for chloride.

Question 2: For detection of sulphur, sodium nitroprusside is added to the sodium extract. A violet colour is obtained due to formation of:
(a) Fe(CN)2
(b) K3Fe(CN)5NS
(c) Na4[Fe(CN)5NOS]
(d) Na4Fe(CN)6
Show Answer & Explanation

Answer: (c) Na4[Fe(CN)5NOS]

Explanation:
1. Na2S + Na2[Fe(CN)5NO] → Na4[Fe(CN)5NOS].
2. The product, sodium thionitroprusside, is violet.
3. Answer: (c)

Teacher's Note:
1. Violet with nitroprusside means sulphide.
2. Do not confuse with the Prussian blue test for nitrogen.

Question 3: 0.45 g of an acid (molecular weight 90) is neutralised by 20 mL of 0.5 N caustic potash. The basicity of the acid is:
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. Equivalents of acid = equivalents of KOH = 20 × 0.5/1000 = 0.01.
2. Equivalent weight = 0.45/0.01 = 45.
3. Basicity = molecular weight/equivalent weight = 90/45 = 2.
4. Answer: (b)

Teacher's Note:
1. Basicity = M/E for an acid.
2. Normality-based titration data give equivalent weight directly.

Question 4: The cation that will not be precipitated by H2S in the presence of dilute HCl is:
(a) Pb2+
(b) Cu2+
(c) Co2+
(d) As3+
Show Answer & Explanation

Answer: (c) Co2+

Explanation:
1. Pb2+, Cu2+ and As3+ belong to group II and form sulphides in acidic medium.
2. Co2+ belongs to group IV, whose sulphides need an alkaline medium (NH4OH present).
3. Answer: (c)

Teacher's Note:
1. Group II: dilute HCl + H2S.
2. Group IV: NH4Cl + NH4OH + H2S.

Question 5: A pink coloured salt turns blue on heating. The presence of which cation is most likely?
(a) Co2+
(b) Cu2+
(c) Zn2+
(d) Fe2+
Show Answer & Explanation

Answer: (a) Co2+

Explanation:
1. Hydrated Co2+ salts are pink, and on dehydration they become blue.
2. Cu2+ salts are blue when hydrated, Zn2+ salts are white and Fe2+ salts are pale green.
3. Answer: (a)

Teacher's Note:
1. Cobalt chloride paper is a standard moisture indicator.
2. Learn colours of hydrated ions.

Question 6: In Kjeldahl's method, CuSO4 acts as:
(a) an oxidising agent
(b) a reducing agent
(c) a hydrolysing agent
(d) a catalytic agent
Show Answer & Explanation

Answer: (d) a catalytic agent

Explanation:
1. In Kjeldahl's digestion, nitrogen is converted to (NH4)2SO4 by hot concentrated H2SO4.
2. CuSO4 speeds up this digestion, so it is a catalyst, while K2SO4 raises the boiling point.
3. Answer: (d)

Teacher's Note:
1. Catalyst: CuSO4.
2. Boiling point raiser: K2SO4.

Question 7: In Lassaigne's test, the organic compound is fused with a piece of sodium metal in order to:
(a) increase the ionisation of the compound
(b) decrease the melting point of the compound
(c) increase the reactivity of the compound
(d) convert the covalent compound into a mixture of ionic compounds
Show Answer & Explanation

Answer: (d) convert the covalent compound into a mixture of ionic compounds

Explanation:
1. Fusion with sodium converts covalently bound N, S and halogens into ionic NaCN, Na2S and NaX.
2. These ionic salts dissolve in water and can be tested by usual reactions.
3. Answer: (d)

Teacher's Note:
1. The sodium extract contains ionic salts.
2. It is also called the sodium fusion extract.

Question 8: A white sodium salt dissolves readily in water to give a solution which is neutral to litmus. When silver nitrate solution is added, a white precipitate is obtained which does not dissolve in dilute HNO3. The anion could be:
(a) CO32−
(b) Cl−
(c) SO42−
(d) S2−
Show Answer & Explanation

Answer: (b) Cl−

Explanation:
1. A neutral solution means a salt of a strong acid and strong base, which rules out Na2CO3 and Na2S (basic).
2. AgCl is a white precipitate insoluble in dilute HNO3, while Ag2SO4 is soluble.
3. Answer: (b)

Teacher's Note:
1. Neutral solution plus AgCl precipitate: chloride.
2. Ag2CO3 dissolves in dilute nitric acid.

Question 9: An aqueous solution of a substance gives a white precipitate on treatment with dil. HCl which dissolves on heating. When H2S is passed through the hot acidic solution, a black precipitate is obtained. The substance is a:
(a) Hg22+ salt
(b) Cu2+ salt
(c) Ag+ salt
(d) Pb2+ salt
Show Answer & Explanation

Answer: (d) Pb2+ salt

Explanation:
1. PbCl2 is a white precipitate that dissolves in hot water.
2. Hot acidic solution with H2S gives black PbS.
3. Answer: (d)

Teacher's Note:
1. PbCl2 is the only common chloride soluble in hot water.
2. AgCl and Hg2Cl2 do not dissolve on heating.

Question 10: A is a lighter phenol and B is an aromatic carboxylic acid. Separation of a mixture of A and B can be carried out easily by using a solution of:
(a) sodium hydroxide
(b) sodium sulphate
(c) calcium chloride
(d) sodium bicarbonate
Show Answer & Explanation

Answer: (d) sodium bicarbonate

Explanation:
1. Carboxylic acids are stronger than carbonic acid and react with NaHCO3 to give soluble sodium salts with CO2.
2. Phenols are weaker than carbonic acid and do not react with NaHCO3.
3. NaOH would dissolve both.
4. Answer: (d)

Teacher's Note:
1. NaHCO3 distinguishes carboxylic acids from phenols.
2. Acid strength decides the reaction.

Question 11: A metal chloride dissolves appreciably in cold water. When placed on a platinum wire in a Bunsen flame no distinctive colour is noticed. The cation would be:
(a) Mg2+
(b) Ba2+
(c) Pb2+
(d) Ca2+
Show Answer & Explanation

Answer: (a) Mg2+

Explanation:
1. Ba2+ gives an apple green flame and Ca2+ a brick red flame.
2. PbCl2 is only sparingly soluble in cold water.
3. MgCl2 is soluble and imparts no flame colour.
4. Answer: (a)

Teacher's Note:
1. Be and Mg give no flame test.
2. Remember flame colours of group II ions.

Question 12: Copper wire test for halogens is known as:
(a) Duma's test
(b) Beilstein's test
(c) Liebig's test
(d) Lassaigne's test
Show Answer & Explanation

Answer: (b) Beilstein's test

Explanation:
1. Organic compounds containing halogens give a green or blue-green flame when heated on a copper wire.
2. This is due to volatile copper halides and is called Beilstein's test.
3. Answer: (b)

Teacher's Note:
1. Duma: nitrogen. Liebig: C and H. Lassaigne: N, S, X.
2. Beilstein's test is not reliable for some nitrogen compounds.

Question 13: In the separation of Cu2+ and Cd2+ in the second group, tetramminecopper(II) sulphate and tetramminecadmium(II) sulphate react with KCN to form cyano complexes. Which pair of complexes and their relative stability enables the separation?
(a) K3[Cu(CN)4] more stable and K2[Cd(CN)4] less stable
(b) K2[Cu(CN)4] less stable and K2[Cd(CN)4] more stable
(c) K2[Cu(CN)4] more stable and K2[Cd(CN)4] less stable
(d) K3[Cu(CN)4] less stable and K2[Cd(CN)4] more stable
Show Answer & Explanation

Answer: (a) K3[Cu(CN)4] more stable and K2[Cd(CN)4] less stable

Explanation:
1. Copper is reduced by cyanide to Cu(I), giving the very stable K3[Cu(CN)4].
2. Cd forms the less stable K2[Cd(CN)4], which still releases enough Cd2+ to give yellow CdS with H2S.
3. The copper complex gives no sulphide, so the two ions are separated.
4. Answer: (a)

Teacher's Note:
1. Cu(I) cyanide complex is extremely stable.
2. The charge on the Cu complex is 3−.

Question 14: Sodium carbonate cannot be used in place of (NH4)2CO3 for the identification of Ca2+, Ba2+ and Sr2+ ions (group V) during mixture analysis because:
(a) Mg2+ ions will also be precipitated
(b) concentration of CO32− is very low
(c) sodium ions will react with acid radicals
(d) Na+ ions will interfere with the detection of Ca2+, Ba2+, Sr2+ ions
Show Answer & Explanation

Answer: (a) Mg2+ ions will also be precipitated

Explanation:
1. Na2CO3 is completely ionised and gives a high CO32− concentration.
2. This exceeds the solubility product of MgCO3 as well, so Mg2+ would also precipitate and group V separation would fail.
3. (NH4)2CO3 in the presence of NH4Cl gives a lower carbonate concentration.
4. Answer: (a)

Teacher's Note:
1. Control of CO32− concentration matters in precipitation.
2. Mg2+ belongs to group VI.

Question 15: A solution containing As3+, Cd2+, Ni2+ and Zn2+ is made alkaline with dilute NH4OH and treated with H2S. The precipitate obtained will consist of:
(a) As2S3 and CdS
(b) CdS, NiS and ZnS
(c) NiS and ZnS
(d) sulphide of all ions
Show Answer & Explanation

Answer: (d) sulphide of all ions

Explanation:
1. NH4OH removes H+ from H2S as water, so sulphide ion concentration rises sharply.
2. The ionic product then exceeds the solubility product of group II sulphides (As2S3, CdS) as well as group IV sulphides (NiS, ZnS).
3. So all four sulphides precipitate.
4. Answer: (d)

Teacher's Note:
1. Acidic medium suppresses S2−; alkaline medium raises it.
2. Group separation depends on S2− concentration.

Question 16: If 0.2 g of an organic compound containing carbon, hydrogen and oxygen on combustion yielded 0.147 g carbon dioxide and 0.12 g water, the percentage of oxygen in the compound is:
(a) 73.29%
(b) 78.45%
(c) 83.23%
(d) 89.50%
Show Answer & Explanation

Answer: (a) 73.29%

Explanation:
1. % C = (12/44) × (0.147/0.2) × 100 = 20.05%.
2. % H = (2/18) × (0.12/0.2) × 100 = 6.67%.
3. % O = 100 − (20.05 + 6.67) = 73.29%.
4. Answer: (a)

Teacher's Note:
1. Oxygen is found by difference.
2. Use 12/44 for C in CO2 and 2/18 for H in H2O.

Question 17: Which of the following statements is incorrect?
(a) Fe2+ ion also gives blood red colour with SCN− ion
(b) Fe3+ ion also gives blood red colour with SCN− ion
(c) On passing H2S through Na2ZnO2 solution a white precipitate of ZnS is formed
(d) Cupric ion reacts with excess of ammonia to give deep blue colour of [Cu(NH3)4]2+ ion
Show Answer & Explanation

Answer: (a) Fe2+ ion also gives blood red colour with SCN− ion

Explanation:
1. Only Fe3+ forms the blood red [Fe(SCN)]2+ with thiocyanate; Fe2+ does not, so statement (a) is incorrect.
2. The other three statements are correct.
3. Answer: (a)

Teacher's Note:
1. Fe2+ gives blue with K3[Fe(CN)6], Fe3+ gives red with SCN−.
2. Read the question: it asks for the incorrect statement.

Question 18: An aqueous solution of a colourless metal sulphate M gives a white precipitate with NH4OH. This is soluble in excess of NH4OH. On passing H2S through this solution a white precipitate is formed. The metal M in the salt is:
(a) Ca
(b) Ba
(c) Al
(d) Zn
Show Answer & Explanation

Answer: (d) Zn

Explanation:
1. Zn(OH)2 is a white precipitate that dissolves in excess ammonia to give [Zn(NH3)4]2+.
2. H2S then gives white ZnS.
3. Ca and Ba do not precipitate with NH4OH; Al(OH)3 does not dissolve in excess and gives no sulphide.
4. Answer: (d)

Teacher's Note:
1. Zn(OH)2 is amphoteric and forms ammine complexes.
2. ZnS is the only common white sulphide in this group.

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Salt Analysis Objective Questions & Solutions for JEE Chemistry

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