JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 31

Chemistry Objective Questions and Answers: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

Question 1: The volume of water that must be added to a mixture of 250 mL of 0.6 M HCl and 750 mL of 0.2 M HCl to obtain a 0.25 M solution is
(a) 200 mL
(b) 750 mL
(c) 300 mL
(d) 100 mL
Show Answer & Explanation

Answer: (a) 200 mL

Explanation:
1. Moles of HCl = 0.25 × 0.6 + 0.75 × 0.2 = 0.15 + 0.15 = 0.30 mol.
2. Final volume for 0.25 M = 0.30 / 0.25 = 1.2 L.
3. Water added = 1200 - 1000 = 200 mL.

Teacher's Note:
1. Total moles stay constant on dilution.
2. Final volume = moles / target molarity.

Question 2: The molarity of Cl- in an aqueous solution that is 2% (w/v) NaCl, 4% (w/v) CaCl2 and 6% (w/v) NH4Cl is (Na 23, Ca 40, N 14, Cl 35.5)
(a) 0.342 M
(b) 0.721 M
(c) 2.18 M
(d) 1.12 M
Show Answer & Explanation

Answer: (c) 2.18 M

Explanation:
1. Per 100 mL: NaCl 2/58.5 = 0.0342 mol; CaCl2 4/111 = 0.0360 mol (2 Cl- each = 0.0721); NH4Cl 6/53.5 = 0.1121 mol.
2. Total Cl- = 0.0342 + 0.0721 + 0.1121 = 0.2184 mol per 100 mL.
3. Molarity = 0.2184 / 0.1 L = 2.18 M.

Teacher's Note:
1. Add the chloride from every salt, counting 2 per CaCl2.
2. Per 100 mL, multiply by 10 for per litre.

Question 3: 100 mL of 2 M Na2SO4 is mixed with 100 mL of 3 M NaCl and 200 mL of 1 M CaCl2. The ratio of the concentration of cations to that of anions is
(a) 2
(b) 1
(c) 1.5
(d) 1/2
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. Cations (mmol): Na from Na2SO4 400, Na from NaCl 300, Ca 200; total 900.
2. Anions (mmol): SO42- 200, Cl- from NaCl 300, Cl- from CaCl2 400; total 900.
3. Ratio = 900 / 900 = 1 (the solution is neutral).

Teacher's Note:
1. Count moles of ions, not charges.
2. Equal moles of cations and anions here is a coincidence of the numbers chosen.

Question 4: A sample of calcium phosphate, Ca3(PO4)2, contains 8 mol of oxygen atoms. The number of moles of calcium atoms in the sample is
(a) 1.5
(b) 8
(c) 4
(d) 3
Show Answer & Explanation

Answer: (d) 3

Explanation:
1. One formula unit of Ca3(PO4)2 has 8 O atoms and 3 Ca atoms.
2. 8 mol O atoms corresponds to 1 mol of the compound, which contains 3 mol Ca.

Teacher's Note:
1. Relate atoms through the formula, not through mass.
2. Ratio Ca : O = 3 : 8.

Question 5: 64 g of an organic compound contains 24 g carbon, 8 g hydrogen and the rest oxygen. Its empirical formula is
(a) CH4O
(b) CH2O
(c) C2H4O
(d) none of these
Show Answer & Explanation

Answer: (a) CH4O

Explanation:
1. Mass of O = 64 - 24 - 8 = 32 g.
2. Moles: C = 2, H = 8, O = 2, ratio 1 : 4 : 1.
3. Empirical formula = CH4O.

Teacher's Note:
1. Subtract to find the missing element.
2. Divide the moles by the smallest.

Question 6: The energy needed by an astronaut for one hour is supplied by burning 34 g of sucrose (C12H22O11) in the body. The mass of oxygen needed for one day (24 h) is about (C = 12, H = 1, O = 16)
(a) 9.162 g
(b) 916.2 g
(c) 91.62 g
(d) 8.162 g
Show Answer & Explanation

Answer: (b) 916.2 g

Explanation:
1. Sucrose per day = 34 × 24 = 816 g; moles = 816 / 342 = 2.386.
2. C12H22O11 + 12O2 → 12CO2 + 11H2O, so O2 = 12 × 2.386 = 28.63 mol.
3. Mass = 28.63 × 32 = 916.2 g.

Teacher's Note:
1. Write and balance the combustion equation first.
2. Scale the hourly requirement to 24 hours.

Question 7: A mixture of 10 mol SO2 and 15 mol O2 is passed over a catalyst and 8 mol of SO3 is formed. How many moles of SO2 and O2 did not combine?
(a) 8 mol SO2, 4 mol O2
(b) 3 mol SO2, 11.5 mol O2
(c) 2 mol SO2, 11 mol O2
(d) 2 mol SO2, 4 mol O2
Show Answer & Explanation

Answer: (c) 2 mol SO2, 11 mol O2

Explanation:
1. 2SO2 + O2 → 2SO3: forming 8 mol SO3 uses 8 mol SO2 and 4 mol O2.
2. Left over: SO2 = 10 - 8 = 2 mol; O2 = 15 - 4 = 11 mol.

Teacher's Note:
1. Use the mole ratio to find what was consumed.
2. Subtract from the initial amount.

Question 8: A piece of iron gains 10% of its mass because of partial rusting to Fe2O3. The percentage of the total iron that has rusted is (Fe = 56, O = 16)
(a) 23
(b) 25.67
(c) 13
(d) 23.3
Show Answer & Explanation

Answer: (d) 23.3

Explanation:
1. Take 100 g of iron; the gain of 10 g is oxygen = 0.625 mol O atoms.
2. Fe2O3 has Fe : O = 2 : 3, so Fe rusted = 0.625 × 2/3 = 0.4167 mol = 23.33 g.
3. Percentage rusted = 23.3%.

Teacher's Note:
1. The mass gained equals the mass of oxygen combined.
2. Use the ratio of atoms in Fe2O3.

Question 9: Polyethene is made from calcium carbide: CaC2 + H2O → Ca(OH)2 + C2H2; C2H2 + H2 → C2H4; nC2H4 → (CH2CH2)n. The mass of polyethene obtainable from 64.0 kg of CaC2 is
(a) 21 kg
(b) 14 kg
(c) 28 kg
(d) 42 kg
Show Answer & Explanation

Answer: (c) 28 kg

Explanation:
1. Moles of CaC2 = 64 000 / 64 = 1000 mol.
2. Each mole of CaC2 gives one mole of C2H2, then one mole of C2H4 (28 g).
3. Mass of polymer = 1000 × 28 g = 28 kg.

Teacher's Note:
1. All steps are 1 : 1 so atom counts carry through.
2. Polymer mass equals the mass of monomer used.

Question 10: x grams of carbon are heated with y grams of oxygen in a closed vessel and no solid residue is left. Which statement is correct?
(a) y/x must be greater than or equal to 2.67
(b) y/x must be less than or equal to 1.33
(c) y/x must be greater than or equal to 1.33
(d) y/x must lie between 1.33 and 2.67
Show Answer & Explanation

Answer: (c) y/x must be greater than or equal to 1.33

Explanation:
1. No solid residue means all the carbon is gone, forming CO or CO2.
2. Minimum oxygen: C + 1/2 O2 → CO gives y/x = 16/12 = 1.33.
3. Any larger ratio (up to complete conversion to CO2 at 2.67 and beyond) still leaves no carbon, so y/x is at least 1.33.

Teacher's Note:
1. Test the extreme products CO and CO2.
2. Excess oxygen cannot leave carbon behind.

Question 11: A 12 g mixture of carbon and sulphur is burnt to give CO2 and SO2, with the moles of SO2 half those of CO2. The mass of carbon in the mixture is
(a) 8.74 g
(b) 1.54 g
(c) 4.08 g
(d) 5.14 g
Show Answer & Explanation

Answer: (d) 5.14 g

Explanation:
1. Let C = n mol; then S = n/2 mol.
2. 12n + 32(n/2) = 12, so 28n = 12 and n = 0.4286 mol.
3. Mass of C = 0.4286 × 12 = 5.14 g.

Teacher's Note:
1. Each C gives one CO2 and each S gives one SO2.
2. Set up the mass equation with the mole relation.

Question 12: A solution of glucose is labelled with mole fraction x and molality m at 10 °C. When these are calculated at 24 °C, you will find
(a) mole fraction x/2 and molality m/2
(b) mole fraction 2x and molality 2m
(c) mole fraction x and molality (m ± dm)
(d) mole fraction x and molality m
Show Answer & Explanation

Answer: (d) mole fraction x and molality m

Explanation:
1. Mole fraction and molality are defined by masses or moles, which do not depend on temperature.
2. So both values stay x and m.

Teacher's Note:
1. Molarity changes with temperature because volume changes.
2. Mass-based and mole-based terms are temperature independent.

Question 13: An aqueous solution of ethanol has density 1.025 g mL-1 and is 2 M. Its molality is
(a) 1.79
(b) 2.143
(c) 1.951
(d) none of these
Show Answer & Explanation

Answer: (b) 2.143

Explanation:
1. 1 L of solution weighs 1025 g and contains 2 mol ethanol = 92 g.
2. Mass of water = 1025 - 92 = 933 g.
3. Molality = 2 / 0.933 = 2.143 m.

Teacher's Note:
1. Find solvent mass by subtraction.
2. Take 1 L as the basis.

Question 14: The mole fraction of ethyl alcohol in aqueous ethyl alcohol is 0.25. The percentage of ethyl alcohol by weight is
(a) 46%
(b) 25%
(c) 54%
(d) 75%
Show Answer & Explanation

Answer: (a) 46%

Explanation:
1. Take 1 mol of solution: 0.25 mol ethanol = 11.5 g; 0.75 mol water = 13.5 g.
2. Mass % = 11.5 / 25 × 100 = 46%.

Teacher's Note:
1. Convert moles to masses before computing mass percent.
2. Mole fraction and mass fraction are different.

Question 15: 4 g of SO3 is dissolved in 100 mL of a sulphuric acid solution containing 80 mass percent H2SO4, density 1.96 g mL-1. The mass percent of H2SO4 in the final solution is (SO3 + H2O → H2SO4; H2SO4 = 98)
(a) 80.8%
(b) 84%
(c) 41.65%
(d) none of these
Show Answer & Explanation

Answer: (a) 80.8%

Explanation:
1. Mass of original solution = 196 g, containing 156.8 g H2SO4 and 39.2 g water.
2. 4 g SO3 = 0.05 mol, which uses 0.9 g water and forms 4.9 g H2SO4.
3. Total H2SO4 = 161.7 g in 200 g of solution = 80.85%, about 80.8%.

Teacher's Note:
1. Total mass after adding SO3 is 196 + 4 = 200 g.
2. Account for the extra H2SO4 formed from SO3 and water.

Question 16: How many gram ions of SO42- are present in 1.25 mol of K2SO4·Al2(SO4)3·24H2O?
(a) 5
(b) 10
(c) 1.25
(d) 2.5
Show Answer & Explanation

Answer: (a) 5

Explanation:
1. One formula unit contains 1 + 3 = 4 sulphate ions.
2. Gram ions = 1.25 × 4 = 5.

Teacher's Note:
1. Add the sulphate from every part of the double salt.
2. Gram ions of an ion means moles of that ion.

Question 17: A solvent contains C, H, O and S. Its hydrogen content is 7.7% by mass and the mass ratio C : O : S = 3 : 2 : 4. The least possible molar mass of the compound is (C = 12, O = 16, S = 32, H = 1)
(a) 39 g
(b) 156 g
(c) 78 g
(d) 62 g
Show Answer & Explanation

Answer: (c) 78 g

Explanation:
1. In 100 g: the remaining 92.3 g is shared 3 : 2 : 4, so C = 30.77 g, O = 20.51 g, S = 41.03 g.
2. Moles: C = 2.56, H = 7.7, O = 1.28, S = 1.28, giving C : H : O : S = 2 : 6 : 1 : 1.
3. Empirical formula C2H6OS, mass = 24 + 6 + 16 + 32 = 78 g. The least molar mass is this empirical mass.

Teacher's Note:
1. Find the empirical formula from the mass ratio, then its mass.
2. The least possible molar mass equals the empirical formula mass.

Question 18: Teflon is made as XeF6 + (CH2-CH2)n → (CF2-CF2)n + HF + XeF4. The moles of XeF6 needed to make 100 g of teflon are
(a) 2
(b) 6
(c) 4
(d) 1
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. Balance per C2H4 unit: C2H4 + 4XeF6 → C2F4 + 4HF + 4XeF4 (24 F on each side).
2. 100 g of teflon = 100 / 100 = 1 mol of C2F4 units.
3. Moles of XeF6 = 4.

Teacher's Note:
1. Count fluorine atoms on both sides to balance.
2. The repeat unit C2F4 has a molar mass of 100 g.

Question 19: A 3 : 2 molar mixture of FeO and Fe2O3 reacts with oxygen to give a 2 : 3 molar mixture of FeO and Fe2O3. The mass of O2 needed per mole of the initial mixture is (O = 16)
(a) 2 g
(b) 1 g
(c) 8 g
(d) 4 g
Show Answer & Explanation

Answer: (a) 2 g

Explanation:
1. 1 mol initial mixture: 0.6 mol FeO and 0.4 mol Fe2O3; Fe = 0.6 + 0.8 = 1.4 mol.
2. Final mixture: FeO = 2k, Fe2O3 = 3k, with 2k + 6k = 1.4, so k = 0.175; O atoms = 0.35 + 1.575 = 1.925 mol.
3. Initial O atoms = 0.6 + 1.2 = 1.8 mol, so O added = 0.125 mol atoms = 0.0625 mol O2 = 2 g.

Teacher's Note:
1. Conserve iron atoms and track the change in oxygen atoms.
2. Convert O atoms to O2 molecules.

Question 20: In a fluorine disposal plant, F2 + 2NaOH → 1/2 O2 + 2NaF + H2O and 2NaF + CaO + H2O → CaF2 + 2NaOH. If 1900 kg of F2 was fed and 10 000 kg of lime (CaO) was supplied, the percentage utilisation of lime was (F = 19, Ca = 40)
(a) 28%
(b) 56%
(c) 72%
(d) 14%
Show Answer & Explanation

Answer: (a) 28%

Explanation:
1. Moles of F2 = 1 900 000 / 38 = 50 000 mol, giving 100 000 mol NaF.
2. CaO used = 100 000 / 2 = 50 000 mol = 2800 kg.
3. Utilisation = 2800 / 10 000 × 100 = 28%.

Teacher's Note:
1. Add the two steps to see F2 + CaO → CaF2 + 1/2 O2.
2. Utilisation = used / supplied × 100.

Question 21: 112 L of Cl2 at STP is passed into 10 L of 1 M KOH. The yields are: Cl2 + 2KOH → KCl + KClO + H2O 60%; 3KClO → 2KCl + KClO3 50%; 4KClO3 → 3KClO4 + KCl 80%. The total moles of KCl produced, rounded to the nearest whole number, is
(a) 6
(b) 4
(c) 3
(d) 5
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. Step 1: Cl2 = 5 mol, KOH = 10 mol; 60% yield gives KCl = KClO = 3 mol.
2. Step 2: 50% of 3 mol KClO reacts, i.e. 1.5 mol, giving 1.0 mol KCl and 0.5 mol KClO3.
3. Step 3: 80% of 0.5 mol KClO3 reacts, giving 0.4 × 1/4 = 0.1 mol KCl.
4. Total KCl = 3 + 1 + 0.1 = 4.1, about 4 mol.

Teacher's Note:
1. Apply each percentage yield in turn.
2. Add KCl from all three steps.

Question 22: When 1 mol of A reacts with 1/2 mol B2 to form AB, 100 kcal is released; when 1 mol of A reacts with 2 mol B2 to form AB4, 200 kcal is released. When 1 mol of A reacts completely with excess B2 to form both AB and AB4, 140 kcal is released. The number of moles of B2 used is
(a) 2.0
(b) 0.6
(c) 1.4
(d) 1.1
Show Answer & Explanation

Answer: (d) 1.1

Explanation:
1. Let x mol A form AB: 100x + 200(1 - x) = 140, so x = 0.6 and AB4 = 0.4 mol.
2. B2 used = 0.6 × 1/2 + 0.4 × 2 = 0.3 + 0.8 = 1.1 mol.

Teacher's Note:
1. Write the energy balance with two unknown fractions of A.
2. Add B2 consumed in both reactions.

Question 23: 92 g of a mixture of CaCO3 and MgCO3 is heated strongly, and the residue (CaO + MgO) weighs 48 g (Ca = 40, Mg = 24). The mass of MgCO3 in the mixture is
(a) 38 g
(b) 46 g
(c) 42 g
(d) 50 g
Show Answer & Explanation

Answer: (c) 42 g

Explanation:
1. Let CaCO3 = a mol and MgCO3 = b mol: 100a + 84b = 92 and 56a + 40b = 48.
2. From the second, a = (48 - 40b)/56; substituting gives b = 0.5 and a = 0.5.
3. Mass of MgCO3 = 0.5 × 84 = 42 g.

Teacher's Note:
1. The CO2 lost is 92 - 48 = 44 g, equal to 1 mol in total.
2. Both carbonates give one CO2 per formula unit, so a + b = 1 mol.

Question 24: Cl2 is passed into a basic solution containing 0.1 mol of S2O32- and 3 mol of OH-, forming SO42- and Cl- (Cl2 is in large excess and no other reaction occurs). The moles of OH- left are
(a) 3
(b) 1
(c) 2
(d) 2.5
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. Balanced equation: S2O32- + 4Cl2 + 10OH- → 2SO42- + 8Cl- + 5H2O.
2. 0.1 mol S2O32- consumes 1.0 mol OH-.
3. OH- left = 3 - 1 = 2 mol.

Teacher's Note:
1. Each S in thiosulphate goes from +2 to +6, so 8 electrons are lost per ion.
2. Balance the charge with hydroxide in basic solution.

Question 25: The mass of water that must be added to 16 g of methanol so that the mole fraction of methanol becomes 0.25 is (C = 12, O = 16)
(a) 9 g
(b) 54 g
(c) 18 g
(d) 27 g
Show Answer & Explanation

Answer: (d) 27 g

Explanation:
1. Moles of methanol = 16/32 = 0.5.
2. x = 0.25 means total moles = 2, so water = 1.5 mol.
3. Mass of water = 1.5 × 18 = 27 g.

Teacher's Note:
1. Mole fraction = moles of solute / total moles.
2. Convert water moles to grams.

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