JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 25

Answer: (c) 93.3 °C

Explanation:
1. Use \(C = \frac{5}{9}(F - 32)\).
2. \(C = \frac{5}{9}(200 - 32) = \frac{5}{9}\times 168 = 93.3\).
3. So the reading is 93.3 °C.

Teacher's Note:
1. Subtract 32 first, then multiply by 5/9.
2. Check by remembering that 212 °F is 100 °C, so 200 °F must be a little below 100 °C.

Answer: (a) \(12.044 \times 10^{20}\) molecules

Explanation:
1. Moles = M × V(L) = 0.02 × 0.1 = 0.002 mol.
2. Molecules = 0.002 × 6.022 × 10^23 = 1.2044 × 10^21.
3. This equals \(12.044 \times 10^{20}\) molecules.

Teacher's Note:
1. Always change mL to litres before using molarity.
2. Molecules = moles × Avogadro's number.

Answer: (a) 4.7 g

Explanation:
1. Mass = density × volume = 3.12 × 1.5 = 4.68 g.
2. In multiplication the result keeps as many significant figures as the least precise number.
3. 1.5 mL has 2 significant figures, so the answer is 4.7 g.

Teacher's Note:
1. Count the significant figures of each given quantity.
2. The least precise quantity (here 1.5 mL) decides the figures in the product.

Answer: (c) A compound retains the physical properties of its constituent elements

Explanation:
1. A compound is formed by chemical combination in a fixed ratio, so it has its own properties.
2. Sodium chloride, for example, is quite different from sodium metal and chlorine gas.
3. Hence the statement that a compound retains the properties of its constituents is incorrect.

Teacher's Note:
1. A mixture keeps the properties of its components; a compound does not.
2. Compounds can be split only by chemical methods.

Answer: (c) Both \(6.022 \times 10^{23}\) molecules of oxygen and 32 g of oxygen

Explanation:
1. One mole of \(\ce{O2}\) contains \(6.022 \times 10^{23}\) molecules.
2. Its mass is 2 × 16 = 32 g.
3. It contains \(1.2044 \times 10^{24}\) atoms, so options with atoms or 16 g are wrong.

Teacher's Note:
1. Oxygen gas is diatomic, so one mole of molecules is 32 g.
2. Do not confuse molecules with atoms.

Answer: (a) (1) and (2)

Explanation:
1. (1) 20 g NaOH = 0.5 mol in 0.2 L, so 2.5 M.
2. (2) 0.5 mol in 0.2 L = 2.5 M.
3. (3) 1 mol in 0.1 L = 10 M. (4) 20/56 = 0.357 mol in 0.2 L = 1.79 M.
4. Only (1) and (2) match.

Teacher's Note:
1. Convert each statement to molarity.
2. Molar mass of NaOH is 40 g and of KOH is 56 g.

Answer: (c) Mole fraction and mole percent

Explanation:
1. Molality has unit mol/kg and molarity has unit mol/L.
2. Mole fraction is a ratio of moles, and mole percent is that ratio multiplied by 100.
3. Both are dimensionless.

Teacher's Note:
1. Ratios of like quantities have no unit.
2. Concentration terms with a volume or mass in the denominator carry units.

Answer: (d) \(\ce{C7H8}\)

Explanation:
1. Moles of water = 0.72/18 = 0.04, so moles of H = 0.08.
2. Moles of \(\ce{CO2}\) = 3.08/44 = 0.07, so moles of C = 0.07.
3. C : H = 0.07 : 0.08 = 7 : 8, so the empirical formula is \(\ce{C7H8}\).

Teacher's Note:
1. Find moles of C from carbon dioxide and moles of H from water.
2. Do not forget that one water molecule has two hydrogen atoms.

Answer: (b) 67.2 L

Explanation:
1. The reaction is \(\ce{2BCl3 + 3H2 -> 2B + 6HCl}\).
2. 21.6 g of boron = 2 mol, which needs 3 mol of \(\ce{H2}\).
3. Volume at STP = 3 × 22.4 = 67.2 L.

Teacher's Note:
1. Balance the equation first and use the mole ratio.
2. At 273 K and 1 atm one mole of gas occupies 22.4 L.

Answer: (a) the mass of one mole of carbon

Explanation:
1. One mole is defined as \(N_A\) particles.
2. A smaller \(N_A\) means one mole would contain far fewer carbon atoms, so its mass would fall.
3. Ratios of species in equations or compounds do not depend on \(N_A\).

Teacher's Note:
1. A mole is a counting unit, and its size depends on the value of \(N_A\).
2. Ratios between particles stay unchanged.

Answer: (d) 0.01 M

Explanation:
1. Moles of urea = \(6.02 \times 10^{20} / 6.02 \times 10^{23} = 10^{-3}\).
2. Volume = 0.1 L.
3. Molarity = 0.001/0.1 = 0.01 M.

Teacher's Note:
1. Convert molecules to moles by dividing by Avogadro's number.
2. Then divide by volume in litres.

Answer: (b) 4 mol

Explanation:
1. \(\ce{2H2 + O2 -> 2H2O}\). Moles of \(\ce{H2}\) = 5, moles of \(\ce{O2}\) = 2.
2. 5 mol of \(\ce{H2}\) would need 2.5 mol of \(\ce{O2}\), so \(\ce{O2}\) is limiting.
3. 2 mol of \(\ce{O2}\) give 4 mol of water.

Teacher's Note:
1. Identify the limiting reagent first.
2. The limiting reagent decides the amount of product.

Answer: (d) 200 amu

Explanation:
1. Average = 0.90 × 200 + 0.08 × 199 + 0.02 × 202.
2. = 180 + 15.92 + 4.04 = 199.96.
3. This is closest to 200 amu.

Teacher's Note:
1. Weight each isotopic mass by its abundance.
2. The average lies close to the most abundant isotope.

Answer: (a) 15 L of \(\ce{H2}\) gas at STP

Explanation:
1. Moles: (a) 15/22.4 = 0.670; (b) 5/22.4 = 0.223.
2. (c) 0.5/2 = 0.25; (d) 10/32 = 0.3125.
3. The largest number of moles is in 15 L of \(\ce{H2}\).

Teacher's Note:
1. Number of molecules is proportional to moles.
2. Convert gas volumes at STP using 22.4 L per mole.

Answer: (d) 3, 3 and 3

Explanation:
1. Leading zeros are never significant.
2. All three numbers contain only the digits 1, 6 and 1 as significant digits.
3. So each has 3 significant figures.

Teacher's Note:
1. Zeros that only locate the decimal point are not significant.
2. Moving the decimal point does not change the number of significant figures.

Answer: (a) all the oxygen will be consumed

Explanation:
1. 1 mol of \(\ce{NH3}\) needs 1.25 mol of \(\ce{O2}\), but only 1 mol is present.
2. So \(\ce{O2}\) is the limiting reagent and is fully used up.
3. NO formed = 0.8 mol and \(\ce{H2O}\) formed = 1.2 mol, so (b) and (c) are wrong.

Teacher's Note:
1. The reagent that runs out first is the limiting reagent.
2. Use the mole ratio 4 : 5 for ammonia and oxygen.

Answer: (a) 6.87

Explanation:
1. Molar mass of \(\ce{CCl4}\) = 12 + 4 × 35.5 = 154 g/mol.
2. Density = M/22.4 = 154/22.4.
3. = 6.875 g/L, nearest to 6.87.

Teacher's Note:
1. Gas density at STP equals molar mass divided by 22.4 L.
2. Calculate the molar mass carefully.

Answer: (a) 1

Explanation:
1. Moles of \(\ce{Mg^{2+}}\) = 12 mg / 24 = 0.5 mmol.
2. Equivalents = 0.5 × 2 = 1 milli-equivalent.
3. Washing soda needed equals the equivalents of \(\ce{Mg^{2+}}\), so 1 meq.

Teacher's Note:
1. Equivalents = moles × charge for an ion.
2. Substances react in equal numbers of equivalents.

Answer: (c) 4

Explanation:
1. Moles of \(\ce{CH3OH}\) = 64/32 = 2; moles of \(\ce{O2}\) = 44.8/22.4 = 2.
2. 2 mol methanol needs 3 mol \(\ce{O2}\), so \(\ce{O2}\) is limiting.
3. \(\ce{CO2}\) = 2 × (2/3) = 4/3 mol, so x = 4.

Teacher's Note:
1. Find the limiting reagent before computing the product.
2. Express the answer in the form requested.

Answer: (b) 2

Explanation:
1. \(\ce{Al(OH)3 + 3HCl -> AlCl3 + 3H2O}\). Moles of HCl = 0.080 × 3.0 = 0.24.
2. Moles of \(\ce{Al(OH)3}\) = 0.08, mass = 0.08 × 78 = 6.24 g.
3. 6.24 = 624 × 10^-2, so x = 2.

Teacher's Note:
1. Balance the equation to get the 1 : 3 mole ratio.
2. Molar mass of \(\ce{Al(OH)3}\) is 78 g.

Answer: (b) 20%

Explanation:
1. \(\ce{6Fe2O3 -> 4Fe3O4 + O2}\). Mass lost = 0.08 g of \(\ce{O2}\) = 0.0025 mol.
2. \(\ce{Fe2O3}\) = 6 × 0.0025 = 0.015 mol = 2.4 g.
3. \(\ce{SiO2}\) = 0.6 g, which is 20% of 3 g.

Teacher's Note:
1. The loss in mass is entirely oxygen gas.
2. Use the balanced equation to link oxygen to \(\ce{Fe2O3}\).

Answer: (c) 24

Explanation:
1. (16a + 28b)/(a + b) = 20 gives 8b = 4a, so a = 2b.
2. For the ratio b : a: (16b + 28a)/(a + b) = (16 + 56)b/3b.
3. = 24.

Teacher's Note:
1. Mean molar mass is the mole-fraction weighted average.
2. First find a/b from the given mean.

Answer: (b) 56 g

Explanation:
1. In X, 1 g of A combines with 3 g of B.
2. In Y, 1 g of A combines with 3 × 7/3 = 7 g of B.
3. For 8 g of A, B = 8 × 7 = 56 g.

Teacher's Note:
1. Reduce both compounds to the same mass of A.
2. This is a direct application of the law of multiple proportions.

Answer: (b) 10

Explanation:
1. Moles: Fe = 7/56 = 0.125, O = 2.5/16 = 0.15625.
2. With x mol FeO and y mol \(\ce{Fe2O3}\): x + 2y = 0.125 and x + 3y = 0.15625, so y = 0.03125 and x = 0.0625.
3. Masses: FeO = 4.5 g, \(\ce{Fe2O3}\) = 5 g, so the ratio is 9 : 10 and b = 10.

Teacher's Note:
1. Write one balance each for iron and for oxygen.
2. Convert the mole results to masses to get the ratio.

Answer: (a) 1

Explanation:
1. Sulphate from \(\ce{Na2SO4}\) = 0.1 mol.
2. \(\ce{Al2(SO4)3}\): 17.1 g/342 = 0.05 mol, giving 0.15 mol sulphate; total = 0.25 mol.
3. Total volume = 600 mL × 5 = 3 L, so molarity = 0.25/3 = 1/12 M and x = 1.

Teacher's Note:
1. Add the moles of the common ion from both salts.
2. Remember that dilution to 5 times means total volume becomes 5 times the mixed volume.