JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 15

Answer: (b) 54.1%

Explanation:
1. Fe in 0.541 g of \( \ce{Fe2O3} \) = 0.541 x 112/160 = 0.3787 g.
2. Percentage = 0.3787/0.700 x 100 = 54.1%.

Teacher's Note:
1. Iron is conserved from the ore to the oxide. Each \( \ce{Fe2O3} \) holds 112 g of iron per 160 g.

Answer: (a) 10%

Explanation:
1. Oxygen gained = 0.90 g = 0.05625 mol.
2. Cu oxidised = 0.05625 mol = 3.6 g.
3. Unoxidised Cu = 0.4 g, which is 10%.

Teacher's Note:
1. The gain in mass equals the oxygen combined, which gives the copper converted to oxide.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. A/2 = 3/2 = 1.5 and B/3 = 4/3 = 1.33, so B is limiting.
2. C formed = 4/3 mol, so both statements are true.
3. The reason explains why the amount of C is fixed by B.

Teacher's Note:
1. Divide each amount by its coefficient. The smaller value is the limiting reagent and fixes the amount of product.

Answer: (d) Assertion is false but the Reason is true.

Explanation:
1. Total mass is conserved in every reaction, so the Reason is true.
2. Total moles generally change (for example, \( \ce{N2 + 3H2 -> 2NH3} \) goes from 4 mol to 2 mol), so the Assertion is false.

Teacher's Note:
1. Mass is conserved but moles are not, unless the number of moles of reactants and products happens to be equal.

Answer: (c) The mass of impurity is 3 g and the metal is Li

Explanation:
1. n of \( \ce{CO2} \) = (1 x 12.315)/(0.0821 x 300) = 0.5 mol, so 0.5 mol of carbonate.
2. \( \ce{Li2CO3} \) (74 g/mol): 37 g, so impurity = 3 g. This fits.
3. \( \ce{BeCO3} \) (69 g/mol): 34.5 g, so impurity 5.5 g, not 1 g or 5 g. \( \ce{MgCO3} \) (84 g/mol): 42 g, more than 40 g, impossible.

Teacher's Note:
1. Work out the moles of carbonate, then test each metal. A carbonate heavier than the whole sample is impossible.

Answer: (c) 1, 2 and 4 only

Explanation:
1. \( \ce{H2SO4} \) gives 2 mol of acidic H+, so it needs 2 mol of OH- equivalents.
2. Ammonia 2 mol: yes. \( \ce{Ba(OH)2} \) 1 mol gives 2 mol OH-: yes. KOH 2 mol: yes.
3. \( \ce{Ca(OH)2} \) 0.5 mol gives only 1 mol OH-: no.

Teacher's Note:
1. Compare equivalents: acidity of the base times its moles must equal 2.

Answer: (a) 2, 3 and 4 only

Explanation:
1. Mg = 0.5 mol, \( \ce{O2} \) = 1 mol. \( \ce{2Mg + O2 -> 2MgO} \) needs 0.25 mol \( \ce{O2} \), so Mg is completely used. Statement 1 is wrong.
2. \( \ce{O2} \) left = 0.75 mol. MgO = 0.5 mol = 20 g. Total mass = 44 g (conservation).
3. Statements 2, 3 and 4 are correct.

Teacher's Note:
1. Find the limiting reagent. Total mass of a closed system never changes.

Answer: (b) 1, 2 and 3 only

Explanation:
1. \( \ce{CaCO3} \) = 0.5 mol, \( \ce{H3PO4} \) = 0.7 mol. 0.5/3 = 0.1667 and 0.7/2 = 0.35, so \( \ce{CaCO3} \) is limiting.
2. \( \ce{Ca3(PO4)2} \) = 0.1667 mol x 310 = 51.67 g. \( \ce{CO2} \) = 0.5 mol.
3. \( \ce{H3PO4} \) used = 0.333 mol, left = 0.367 mol = 35.93 g. So 1, 2 and 3 are correct; 4 is not.

Teacher's Note:
1. Find the limiting reagent and compute every quantity from it.

Answer: (b) 1 and 4 only

Explanation:
1. Let each reaction use a mol of A. B = a + a/2 = 1.5a. C = a/2.
2. So B/C = 3 always, and B is greater than C.
3. If C = 2 mol, B = 6 mol. So statements 1 and 4 are correct.

Teacher's Note:
1. Find B and C in terms of one quantity and take their ratio.

Answer: (a) A-R, B-Q, C-P

Explanation:
1. Moles in daily dose = 0.0176/176 = \( 10^{-4} \) mol (C-P).
2. O atoms = 6 x \( 10^{-4} \) x \( 6\times10^{23} \) = \( 3.6\times10^{20} \) (A-R).
3. Moles in 1 g = 1/176 = \( 5.68\times10^{-3} \) (B-Q).

Teacher's Note:
1. Convert between mass, moles and atoms using the molar mass and Avogadro number.

Answer: (b) 420 kg

Explanation:
1. Balanced: 3FeBr2 + Br2 gives Fe3Br8; Fe3Br8 + 4Na2CO3 gives 8NaBr + 4CO2 + Fe3O4.
2. 3 Fe gives 1 Fe3Br8 which gives 8 NaBr. NaBr = 2060 kg/103 = 20 kmol.
3. Fe = 20 x 3/8 = 7.5 kmol = 420 kg.

Teacher's Note:
1. Track iron through the three steps: three Fe atoms end up making eight NaBr.

Answer: (d) \( 10^3 \) kg

Explanation:
1. At 100% yield the iron needed is 420 kg.
2. Overall yield = 0.6 x 0.7 = 0.42.
3. Iron required = 420/0.42 = 1000 kg.

Teacher's Note:
1. Multiply the step yields to get the overall yield and divide the theoretical amount by it.

Answer: (c) 10

Explanation:
1. NaBr formed = 2060/103 = 20 mol.
2. In reaction (iii), 8 NaBr form together with 4 \( \ce{CO2} \), so \( \ce{CO2} \) = 20/2 = 10 mol.
3. The yield only changes how much reactant is used. The products form in fixed ratio, so \( \ce{CO2} \) = 10 mol.

Teacher's Note:
1. Products of one reaction always appear in their coefficient ratio, whatever the yield.

Answer: (c) 250

Explanation:
1. Theoretical requirement = 32.2 kg.
2. Overall yield = 0.6 x 0.6667 x 0.322 = 0.1288.
3. Mass needed = 32.2/0.1288 = 250 kg.

Teacher's Note:
1. Divide the theoretical mass by the product of the fractional yields.

Answer: (c) HF

Explanation:
1. Mass of F in gas = 0.95 x 0.8 = 0.76 g = 0.04 mol. Mass of H = 0.04 g = 0.04 mol.
2. F : H = 1 : 1, so the formula is HF.

Teacher's Note:
1. Convert the mass of each element to moles and take the simplest ratio.

Answer: (a) S - 2 - iv

Explanation:
1. Urea \( \ce{CO(NH2)2} \) (60 g/mol): %H = 20/3, %N = 46.7, atoms of C = 4 N_A, total atoms 32 N_A = 2 x 16 N_A (H atoms). Acetic acid \( \ce{C2H4O2} \) (2 mol): %C = 40, %H = 20/3, %O = 160/3, C and O atoms = 8 N_A, total atoms 16 N_A, H atoms 8 N_A. Glucose \( \ce{C6H12O6} \) (2/3 mol): %C = 40, %H = 20/3, %O = 160/3, C = 4 N_A, C and O = 8 N_A, total atoms 16 N_A, H atoms 8 N_A. The hydrocarbon has 80% C: C = 4 mol, H = 12 mol, total atoms 16 N_A.
2. S-2-iv is correct: glucose has %H = 20/3 and 16 N_A total atoms, which is twice its 8 N_A hydrogen atoms.
3. P-1-i and P-1-ii fail since the hydrocarbon has 80% C. Q-2-iii fails since urea has 32 N_A atoms.

Teacher's Note:
1. Evaluate each match against the actual formula before choosing.

Answer: (a) P - 4 - iii

Explanation:
1. Urea \( \ce{CO(NH2)2} \) (60 g/mol): %H = 20/3, %N = 46.7, atoms of C = 4 N_A, total atoms 32 N_A = 2 x 16 N_A (H atoms). Acetic acid \( \ce{C2H4O2} \) (2 mol): %C = 40, %H = 20/3, %O = 160/3, C and O atoms = 8 N_A, total atoms 16 N_A, H atoms 8 N_A. Glucose \( \ce{C6H12O6} \) (2/3 mol): %C = 40, %H = 20/3, %O = 160/3, C = 4 N_A, C and O = 8 N_A, total atoms 16 N_A, H atoms 8 N_A. The hydrocarbon has 80% C: C = 4 mol, H = 12 mol, total atoms 16 N_A.
2. P-4-iii is incorrect because the hydrocarbon contains no nitrogen. The other three are correct.

Teacher's Note:
1. An incorrect match often pairs an element with a compound that does not contain it.

Answer: (b) \( \dfrac{1}{9.108\times6.023}\times10^{8} \)

Explanation:
1. Number of electrons in 1 kg = \( 1/9.108\times10^{-31} = (1/9.108)\times10^{31} \).
2. Moles = number/\( N_A \) = \( \dfrac{1}{9.108}\times10^{31}\times\dfrac{1}{6.023\times10^{23}} = \dfrac{1}{9.108\times6.023}\times10^{8} \).

Teacher's Note:
1. Divide the number of particles by Avogadro number to get moles.

Answer: (c) 55.95

Explanation:
1. Average = (54 x 5 + 56 x 90 + 57 x 5)/100.
2. = (270 + 5040 + 285)/100 = 55.95.

Teacher's Note:
1. Average atomic mass is the abundance-weighted mean of the isotopic masses.

Answer: (c) 2992

Explanation:
1. \( \ce{(NH4)2SO4} \) = 1584/132 = 12 mol, giving 24 mol \( \ce{NH3} \) and 12 mol gypsum (172 g/mol) = 2064 g.
2. \( \ce{NiCl2.6H2O} \) = 952/238 = 4 mol, which needs 24 mol \( \ce{NH3} \). Both are used completely.
3. Complex \( \ce{[Ni(NH3)6]Cl2} \) = 232 g/mol x 4 = 928 g.
4. Total = 2064 + 928 = 2992 g.

Teacher's Note:
1. Check that the ammonia exactly matches the nickel salt, then add the two product masses.