Q: A mixture of two salts is used to prepare a solution S , which gives the following results: White precipitate s only ← Room temperature Dilute NaOH ( aq ) S aq . &#16

P b 2 + → dil . HCl PbCl 2 (white precipitate) P b 2 + → dil . NaOH Pb OH 2 (white precipitate) Bi 3 + → dil HCl BiCl 3 (water-soluble) Bi 3 + → dil . NaOH Bi OH 3 (white precipitate) Hg 2 + → dil . HCl HgCl 2 (water-soluble) Hg 2 + → dil . NaOH Hg OH 2 (yellow precipitate) Ag + → dil . HCl AgCl (white precipitate) Ag + → dil . NaOH AgOH (brown precipitate) Zn 2 + → dil . HCl ZnCl 2 (water-soluble) Zn

Q: A colorless aqueous solution contains nitrates of two metals, X and Y . When it was added to an aqueous solution of NaCl , a white precipitate was formed. This precipitate was found to be partly solub

X = AgNO 3 , Y = Pb NO 3 2 AgNO 3 X + NaCl → AgCl ( P ) ↓ + NaNO 3 AgCl ( s ) ↓ ( P ) + 2 NH 3 → Ag NH 3 2 + Soluble Cl − AgCl ( P ) + 2 Na 2 S 2 O 3 → Na 3 Ag S 2 O 3 2 Solubel + NaCl Pb NO 3 2 Y + 2 NaCl → PbCl 2 White ppt Q ( Soluble in hot water ) ↓ + 2 NaNO 3 PbCl 2 aq . Q + 2 KI → PbI 2 Yellow ppt + 2 KCl

Q: The green colour produced in the borax bead test of a chromium(III) salt is due to -

Borax bead test Na 2 B 4 O 7 .10H 2 O → Δ Na 2 B 4 O 7 → Δ NaBO 2 + B 2 O 3 C r 2 O 3 + B 2 O 3 → Δ G r e e n c o l o u r e d C r ( B O 2 ) 3

Question 1

Passage: Consider the following reaction sequence in which J , K , L and M are the major products. In sulphur estimation by Carius method, the amount of BaSO_4 formed from 3.79 g of M is ____ g.

Accepted Answer

Step 1: The starting material is m-xylene (1,3-dimethylbenzene). It undergoes Friedel-Crafts acylation with chloroacetyl chloride ( ClCH _2 COCl ) in the presence of anhydrous AlCl _3 to form 2-chloro-1-(2,4-dimethylphenyl)ethanone. Step 2: The chloro ketone reacts with NaI (Finkelstein reaction) to form 2-iodo-1-(2,4-dimethylphenyl)ethanone. Step 3: This iodo compound undergoes Williamson ether synthesis with sodium 3-nitrophenoxide to form compound J . The molecular formula of J is C _ 16 H _

Question 2

The correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions, is LIST - I LIST - II (P) Passing H _2 ~S in the presence of NH _4 OH (1) Cu ^ 2+ (Q)

Accepted Answer

Mn ^ +2 H _2 ~S + NH _4 OH Pink/buff ppt. MnS Ba ^ +2 ( NH _4 )_2 CO _3+ NH _4 OH White ppt. BaCO _3 Al ^ +3 NH _4 Cl + NH _4 OH White ppt. Al ( OH )_3 Cu ^ +2 H _2 ~S + HCl (dil.) Black ppt. CuS

Question 3

A mixture of two salts is used to prepare a solution S , which gives the following results: White &#160; precipitate s &#160; only &#8592; Room &#160; temperature Dilute &#160; NaOH ( aq ) S aq . &#16

Accepted Answer

P b 2 + &#8594; dil . HCl PbCl 2 (white precipitate) P b 2 + &#8594; dil . NaOH Pb OH 2 (white precipitate) Bi 3 + &#8594; dil &#160; HCl BiCl 3 (water-soluble) Bi 3 + &#8594; dil . &#160; NaOH Bi OH 3 (white precipitate) Hg 2 + &#8594; dil . HCl HgCl 2 (water-soluble) Hg 2 + &#8594; dil . &#160; NaOH Hg OH 2 (yellow precipitate) Ag + &#8594; dil . &#160; HCl AgCl (white precipitate) Ag + &#8594; dil . &#160; NaOH AgOH (brown precipitate) Zn 2 + &#8594; dil . &#160; HCl ZnCl 2 (water-soluble) Zn

Question 4

A colorless aqueous solution contains nitrates of two metals, X and Y . When it was added to an aqueous solution of NaCl , a white precipitate was formed. This precipitate was found to be partly solub

Accepted Answer

X = AgNO 3 ,   Y = Pb NO 3 2 AgNO 3 X + NaCl &#8594; AgCl ( P ) &#160; &#8595; + NaNO 3 AgCl &#8201; ( s ) &#8595; ( P ) + 2 NH 3 &#8594; Ag NH 3 2 + Soluble Cl &#8722; AgCl ( P ) + 2 Na 2 S 2 O 3 &#8594; Na 3 Ag S 2 O 3 2 Solubel + NaCl Pb NO 3 2 Y + 2 NaCl &#8594; PbCl 2 White &#8201; ppt &#8201; Q ( Soluble &#8201; in &#8201; hot water ) &#8201;&#160;&#8595; + 2 NaNO 3   PbCl 2 aq . Q + 2 KI &#8594; PbI 2 Yellow &#8201; ppt + 2 KCl

Question 5

The green colour produced in the borax bead test of a chromium(III) salt is due to -

Accepted Answer

Borax bead test Na 2 B 4 O 7 .10H 2 O&#8201; &#8594; &#916; Na 2 B 4 O 7 &#8201; &#8594; &#916; NaBO 2 &#8201; + B 2 O 3 C r 2 O 3 + B 2 O 3 &#8201; &#8594; &#916; G r e e n &#160; c o l o u r e d &#8201; &#8201;&#8201;&#8201;&#8201; C r ( B O 2 ) 3

Practical Chemistry — Previous Year Questions & Solutions