JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 27

Answer: (a) 17 g

Explanation:
1. Mass of one atom = 51/3 = 17 u.
2. The atomic mass is 17 u, so the gram atomic mass is 17 g.

Teacher's Note:
1. GAM is the atomic mass expressed in grams.
2. The unit u is replaced by g.

Answer: (c) 4

Explanation:
1. The mass number of each atom is about 2 (2 nucleons).
2. 2 atoms contain 2 \(\times\) 2 = 4 nucleons.

Teacher's Note:
1. Nucleons = protons + neutrons.
2. Atomic mass in u is approximately the number of nucleons.

Answer: (a) \(^{16}\)O

Explanation:
1. Mass of X\(_2\) = (8/27) \(\times\) 108 = 32 u.
2. So the atomic mass of X is 16, which is oxygen-16.

Teacher's Note:
1. Ratio of masses of single particles equals the ratio of their molar masses.
2. X\(_2\) is diatomic.

Answer: (d) both (1) and (3)

Explanation:
1. Molecular mass of H\(_2\)SO\(_4\) = 98 u, so 10 molecules weigh 980 u.
2. 1 u = \( 1.66 \times 10^{-27} \) kg, so the mass is also \( 980 \times 1.66 \times 10^{-27} \) kg. Hence both (1) and (3) are correct.

Teacher's Note:
1. amu and kg are linked by 1 u = 1.66 \(\times 10^{-27}\) kg.
2. 980 g would be 10 moles of H\(_2\)SO\(_4\).

Answer: (d) all of these

Explanation:
1. All three statements are correct definitions.
2. So the answer is "all of these".

Teacher's Note:
1. A mole is Avogadro's number of particles.
2. GAM and GMM are the masses of one mole of atoms or molecules.

Answer: (a) \( 3.32 \times 10^{-22} \) g

Explanation:
1. Mass = 200 \(\times\) 1.66 \(\times 10^{-24}\) g.
2. This is \( 3.32 \times 10^{-22} \) g.

Teacher's Note:
1. 1 amu = \( 1.66 \times 10^{-24} \) g.
2. Or divide 200 by N\(_A\).

Answer: (c) 96500 C

Explanation:
1. Charge = N\(_A\) \(\times\) e = \( 6.022 \times 10^{23} \times 1.6 \times 10^{-19} \) C.
2. This is about 96500 C (one faraday).

Teacher's Note:
1. This value is called one faraday.
2. It is used in electrolysis.

Answer: (c) \( 1.66 \times 10^{-24} \)

Explanation:
1. Moles = number of atoms / N\(_A\) = 1/(\( 6.023 \times 10^{23} \)).
2. This is \( 1.66 \times 10^{-24} \) mole.

Teacher's Note:
1. Note 1/N\(_A\) equals the mass in grams of 1 amu.
2. Do not confuse moles with atoms.

Answer: (b) \( 48.18 \times 10^{22} \)

Explanation:
1. Atoms = 0.1 \(\times\) 8 \(\times\) \( 6.023 \times 10^{23} \) = 0.8 N\(_A\).
2. This is \( 4.818 \times 10^{23} = 48.18 \times 10^{22} \).

Teacher's Note:
1. S\(_8\) has 8 atoms per molecule.
2. Atoms = moles \(\times\) atomicity \(\times\) N\(_A\).

Answer: (a) 2 mol H\(_2\) molecules = 2 g

Explanation:
1. 2 mol of H\(_2\) has mass 2 \(\times\) 2 = 4 g, not 2 g.
2. The other three are correct: 3 \(\times\) 20 = 60 g, 4 \(\times\) 32 = 128 g, 6 \(\times\) 48 = 288 g.

Teacher's Note:
1. Convert everything to moles first.
2. Molar mass of H\(_2\) is 2 g/mol.

Answer: (a) 1 mg CH\(_4\)

Explanation:
1. Atoms per mg (in units of 10\(^{-3}\) mol \(\times\) atomicity): CH\(_4\): 5/16 = 0.313; N\(_2\): 2/28 = 0.071; Na: 1/23 = 0.043; H\(_2\)O: 3/18 = 0.167.
2. CH\(_4\) has the largest number.

Teacher's Note:
1. Compare (mass/molar mass) \(\times\) atoms per molecule.
2. Light molecules with many atoms win.

Answer: (a) 1, 2 and 3 only

Explanation:
1. (1) is correct by definition.
2. (2) 2 mol C = 24 g. (3) 6 \(\times 6.022 \times 10^{23}\) = \( 3.6 \times 10^{24} \) H atoms.
3. (4) is wrong: one molecule has 18 protons, one mole has 18 N\(_A\).

Teacher's Note:
1. Multi-statement questions need every statement checked.
2. Always distinguish per molecule from per mole.

Answer: (c) \( 32 \times 10^{-4} \) N\(_A\)

Explanation:
1. Moles = 6.2 \(\times 10^{-3}\)/62 = \( 10^{-4} \).
2. Electrons per ion = 7 + 24 + 1 = 32.
3. Total = \( 32 \times 10^{-4} \) N\(_A\).

Teacher's Note:
1. The negative charge adds one electron.
2. Electrons per ion = sum of atomic numbers + charge.

Answer: (a) 24.5 g

Explanation:
1. Moles of O\(_2\) = 48/32 = 1.5; mass of H\(_2\) = 0.5 \(\times\) 2 = 1 g.
2. Total mass = 49 g and total moles = 2.
3. Average molar mass = 49/2 = 24.5 g.

Teacher's Note:
1. Average molar mass = total mass / total moles.
2. Do not average the molar masses directly.

Answer: (d) 36 g H\(_2\)O

Explanation:
1. Moles: 54/108 = 0.5; 28/28 = 1; 36/18 = 2; 46/46 = 1.
2. Water has the largest number of moles.

Teacher's Note:
1. Molecules are proportional to moles.
2. Small molar mass favours more molecules per gram.

Answer: (d) 24.5

Explanation:
1. Mass = 0.25 \(\times\) 98 = 24.5 g.

Teacher's Note:
1. Mass = moles \(\times\) molar mass.
2. Molar mass of H\(_2\)SO\(_4\) is 98 g/mol.

Answer: (a) 1.5 \(\times\) 6.02 \(\times 10^{23}\)

Explanation:
1. Moles of CO\(_2\) = 22/44 = 0.5.
2. Atoms = 0.5 \(\times\) 3 \(\times\) N\(_A\) = 1.5 \(\times\) 6.02 \(\times 10^{23}\).

Teacher's Note:
1. Each CO\(_2\) molecule has 3 atoms.
2. Atoms = moles \(\times\) 3 \(\times\) N\(_A\).

Answer: (b) 0 K

Explanation:
1. From V = nRT/P, V = 0 when T = 0 K.
2. 0 \(^\circ\)C is 273 K and 0 \(^\circ\)F is 255 K, so they are not zero volume points.

Teacher's Note:
1. Absolute zero is 0 K.
2. Real gases liquefy before reaching it.

Answer: (d) 50 K

Explanation:
1. T = PV/(nR) = (3.32 \(\times\) 5)/(4 \(\times\) 0.08314) = 49.9 K, about 50 K.

Teacher's Note:
1. With pressure in bar use R = 0.08314 L bar K\(^{-1}\) mol\(^{-1}\).
2. Match the units of P and R.

Answer: (c) 25.215 atm

Explanation:
1. n = 4/32 + 2/2 = 0.125 + 1 = 1.125 mol.
2. P = nRT/V = 1.125 \(\times\) 0.0821 \(\times\) 273/1 = 25.2 atm.

Teacher's Note:
1. Ideal gas law applies to the mixture.
2. Add the moles of all gases.

Answer: (a) P = 0.5 atm, T = 600 K

Explanation:
1. V is proportional to T/P: 1200, 75, 300 and 500 respectively.
2. The largest value is for P = 0.5 atm, T = 600 K.

Teacher's Note:
1. Volume increases with T and decreases with P.
2. Compare the ratio T/P.

Answer: (b) 0.005 mole

Explanation:
1. Moles = 112/22400 = 0.005.
2. One mole occupies 22400 cm\(^3\) at STP.

Teacher's Note:
1. 22.4 L = 22400 mL at STP.
2. Convert cm\(^3\) correctly.

Answer: (c) CO

Explanation:
1. Moles = 2.24/22.4 = 0.1.
2. Molar mass = 2.8/0.1 = 28 g/mol, which is CO (also N\(_2\)).

Teacher's Note:
1. Find molar mass = mass / moles.
2. Compare with the options.

Answer: (c) Atoms of an element may be different due to the presence of isotopes.

Explanation:
1. Dalton believed all atoms of an element are identical.
2. Isotopes (atoms of the same element with different masses) were discovered later.

Teacher's Note:
1. Isotopes contradict Dalton's second postulate.
2. The other three are his postulates.

Answer: (c) 1/12 of the mass of a C-12 atom

Explanation:
1. By definition 1 amu is one twelfth of the mass of a C-12 atom.
2. Its value is \( 1.66 \times 10^{-27} \) kg, not \( 10^{-23} \) kg.

Teacher's Note:
1. 1 amu is about the mass of one proton.
2. Memorise 1.66 \(\times 10^{-24}\) g.

Answer: (c) \( \frac{1}{4} \times 6 \times 10^{23} \)

Explanation:
1. Moles of He in 1 g = 1/4.
2. Atoms = (1/4) \(\times 6 \times 10^{23}\).

Teacher's Note:
1. He is monatomic.
2. Number of atoms is inversely proportional to atomic mass for equal mass.

Answer: (c) y

Explanation:
1. Moles of A in x g = x/40.
2. Moles of B in 2x g = 2x/80 = x/40.
3. Equal moles means equal atoms, y.

Teacher's Note:
1. Equal moles contain equal numbers of atoms.
2. Compare moles, not masses.