JEE Chemistry Redox Reactions MCQs Set 03

Answer: (b) 35%

Explanation:
1. Let KOH = x g: \( \frac{x}{56} + \frac{4.2-x}{37} = 0.1 \) (equivalent weight of Ca(OH)2 = 37)
2. \( x = 1.47 \) g
3. KOH \( = \frac{1.47}{4.2}\times100 = 35\% \)

Teacher's Note:
1. Equivalents of the two bases add up to the equivalents of acid.
2. Ca(OH)2 is diacidic, so its equivalent weight is half its molar mass.

Answer: (b) 157.9 mL

Explanation:
1. Let each be n mol: \( 106n + 84n = 1 \Rightarrow n = \frac{1}{190} \)
2. Complete reaction: \( 2n + n = 3n = 0.01579 \) mol HCl
3. Volume \( = \frac{0.01579}{0.1} = 0.1579 \) L

Teacher's Note:
1. Na2CO3 consumes 2 mol HCl to reach CO2, NaHCO3 consumes 1 mol.

Answer: (b) 20.7%

Explanation:
1. Free SO3 becomes H2SO4 on dilution (equivalent weight 40 for SO3, 49 for H2SO4)
2. \( \frac{x}{40} + \frac{0.5-x}{49} = 0.0107 \)
3. \( x = 0.1036 \) g
4. \( 20.7\% \)

Teacher's Note:
1. Each SO3 gives one H2SO4, which has two replaceable hydrogens.
2. So equivalent weight of SO3 is 80/2 = 40.

Answer: (b) 25 mL

Explanation:
1. HCl taken = 0.25 mol
2. CaCO3 = 0.1 mol uses 0.2 mol HCl
3. Excess HCl = 0.05 mol
4. KOH needed = 0.05/2 = 0.025 L

Teacher's Note:
1. Boiling removes dissolved CO2, so only excess HCl is titrated.

Answer: (b) 200 mL

Explanation:
1. NaH2PO4 has molar mass 120, so n = 0.1 mol
2. It must lose two H+
3. NaOH needed = 0.2 mol = 200 mL

Teacher's Note:
1. Count the number of H+ removed per formula unit.

Answer: (b) 48.8%

Explanation:
1. HCl used = 40 - 4 = 36 mmol
2. \( 2\left(\frac{x}{100}+\frac{1.64-x}{84}\right) = 0.036 \)
3. \( x = 0.8 \) g
4. \( 48.8\% \)

Teacher's Note:
1. Each carbonate uses two moles of HCl.
2. Use molar masses 100 and 84.

Answer: (c) 83.3%

Explanation:
1. Total HCl = 40 meq
2. In 25 mL, excess = 3.75 meq, so in 100 mL = 15 meq
3. HCl used = 25 meq = 1.25 g CaCO3 (E = 50)
4. 1.25/1.5 = 83.3%

Teacher's Note:
1. Scale the aliquot to the full volume before subtracting.

Answer: (a) 0.424 g and 0.21 g

Explanation:
1. Phenolphthalein end point converts Na2CO3 to NaHCO3: 4 meq, so Na2CO3 = 4 mmol = 0.424 g
2. Total with methyl orange 10.5 meq; Na2CO3 accounts for 8
3. NaHCO3 = 2.5 mmol = 0.21 g

Teacher's Note:
1. The phenolphthalein volume equals half the methyl orange titre of Na2CO3.

Answer: (b) 23.2

Explanation:
1. Second stage (MO - Ph) = 25 - 19.5 = 5.5 mL, which equals the half of Na2CO3 neutralised after phenolphthalein
2. Na2CO3 = 5.5 x 0.995 = 5.47 mmol in 25 mL
3. Mass = 5.47 x 106 = 580 mg per 25 mL
4. = 23.2 g/L

Teacher's Note:
1. Phenolphthalein neutralises NaOH plus half of Na2CO3.
2. The methyl orange stage neutralises the other half.

Answer: (c) 0.06 g

Explanation:
1. NaOH + half Na2CO3 = 17.5 mL; the second stage 2.5 mL equals half Na2CO3
2. NaOH = 15 mL of N/10 = 1.5 meq
3. Mass = 1.5 x 40 = 60 mg = 0.06 g

Teacher's Note:
1. Subtract the methyl orange volume from the phenolphthalein volume.

Answer: (b) 5.04

Explanation:
1. Phenolphthalein: 2 mL x 0.2 N = 0.4 meq = Na2CO3 to NaHCO3
2. Second stage: 2.5 x 0.4 N = 1.0 meq, of which 0.4 is from Na2CO3
3. Original NaHCO3 = 0.6 meq = 0.6 x 84 = 50.4 mg per 10 mL = 5.04 g/L

Teacher's Note:
1. The second stage neutralises all bicarbonate.

Answer: (c) 60.15%

Explanation:
1. Na2CO3 (to bicarbonate) = 30 - 25 = 5 mL; total Na2CO3 = 5 mL x 0.1 = 0.5 mmol per 10 mL (= 5 mmol in 100 mL = 0.530 g)
2. NaOH = (25 - 5) x 0.1 = 2 mmol per 10 mL = 20 mmol = 0.8 g
3. % NaOH = 0.8/1.33 = 60.15%

Teacher's Note:
1. Convert both to mass before taking the percentage.

Answer: (c) +3

Explanation:
1. Sn2+ loses 2 e- per ion: 20 mmol gives 40 mmol e-
2. Ce4+ 40.05 mmol takes about 1 e- each
3. So Ce becomes +3

Teacher's Note:
1. Electrons lost equal electrons gained.

Answer: (b) 0

Explanation:
1. Cr2+ gives 2.552 mmol of e-
2. SeO2 = 0.638 mmol
3. Electrons per Se = 2.552/0.638 = 4
4. Se(+4) gains 4 e- and goes to 0

Teacher's Note:
1. Count electrons transferred per atom.

Answer: (b) 31.7 mL

Explanation:
1. Each formula unit has 4 oxalate units, giving 8 e-
2. Moles = 1/505
3. Eq = 8/505 = 0.01584
4. KMnO4 (5 e-): 0.01584/5 = 3.17 mmol, so 31.7 mL of 0.1 M

Teacher's Note:
1. Each oxalate loses 2 electrons.

Answer: (b) 0.588 N

Explanation:
1. Equivalents of H2O2 = (x/100)/17
2. N x (x/1000) = x/1700
3. N = 1000/1700 = 0.588

Teacher's Note:
1. The x cancels out.

Answer: (c) 337 mL

Explanation:
1. Sn to Sn4+ in the net reaction gives 4 e- per atom
2. Equivalents = 4/118.7 = 0.0337
3. Volume of 0.1 N = 0.0337/0.1 = 0.337 L

Teacher's Note:
1. Tin supplies four electrons per atom.

Answer: (b) 41.5%

Explanation:
1. Equivalents in 20 mL = 0.654 meq
2. Cu = 0.654 x 63.5 = 41.5 mg per 20 mL
3. In 1 L: 50 x 41.5 = 2.077 g
4. 2.077/5 = 41.5%

Teacher's Note:
1. Scale from the aliquot to 1 L.

Answer: (c) 6

Explanation:
1. Fe2O3 = 0.552/160 = 3.45 mmol, Fe2+ = 6.9 mmol
2. In 25 mL: 1.725 mmol e- lost
3. Oxidant = 17 x 0.0167 = 0.284 mmol
4. 1.725/0.284 = 6.07, which is 6

Teacher's Note:
1. Electrons lost by Fe2+ equal electrons gained by the oxidant.

Answer: (b) 0.15 N

Explanation:
1. Fe in the ore = 0.0084x g
2. Equivalents = 0.0084x/56
3. N x x/1000 = 0.0084x/56
4. N = 0.15

Teacher's Note:
1. The x cancels again.

Answer: (c) 90.1%

Explanation:
1. Acid added = 5 meq; back titration used 1.6 meq
2. Acid used = 3.4 meq = 3.4 x 53 = 180.2 mg
3. The 50 mL portion holds 200 mg sample, so purity = 90.1%

Teacher's Note:
1. A back titration measures the excess of acid.