JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 28

Chemistry Objective Questions and Answers: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

Question 1: The number of atoms in 558.5 g of Fe (atomic weight 55.85) is:
(a) \( 558.5 \times 6.023 \times 10^{23} \)
(b) half that in 8 g of He
(c) \( 6.022 \times 10^{22} \)
(d) twice that in 60 g of carbon
Show Answer & Explanation

Answer: (d) twice that in 60 g of carbon

Explanation:
1. 558.5 g Fe = 10 mol.
2. 60 g C = 5 mol, so Fe has twice as many atoms.
3. 8 g He = 2 mol, so Fe has five times as many (not half).

Teacher's Note:
1. Convert every option to moles.
2. Number of atoms = moles \(\times\) N\(_A\).

Question 2: Which of the following has the maximum mass?
(a) 10 mL of water
(b) \( 3.011 \times 10^{23} \) atoms of oxygen
(c) 1 g-atom of C
(d) 1/2 mole of CH\(_4\)
Show Answer & Explanation

Answer: (c) 1 g-atom of C

Explanation:
1. 1 g-atom C = 12 g; 1/2 mol CH\(_4\) = 8 g; 10 mL water = 10 g; 0.5 mol O atoms = 8 g.
2. The maximum is 12 g.

Teacher's Note:
1. Convert each to grams.
2. Density of water is 1 g/mL.

Question 3: The total number of protons, electrons and neutrons in 12 g of \(^{12}_{6}\)C is:
(a) \( 6.022 \times 10^{23} \)
(b) 18
(c) \( 1.084 \times 10^{25} \)
(d) \( 6.022 \times 10^{22} \)
Show Answer & Explanation

Answer: (c) \( 1.084 \times 10^{25} \)

Explanation:
1. Each atom has 6 protons, 6 electrons and 6 neutrons = 18 particles.
2. 12 g = 1 mol, so total = 18 \(\times 6.022 \times 10^{23} = 1.084 \times 10^{25}\).

Teacher's Note:
1. 12 g of C-12 is one mole.
2. Multiply particles per atom by N\(_A\).

Question 4: One mole of element X has a mass 3/10 times the mass of one mole of element Y. One average atom of X has a mass 2 times the mass of one atom of \(^{12}\)C. What is the atomic weight of Y?
(a) 46.67
(b) 80
(c) 40.0
(d) 15.77
Show Answer & Explanation

Answer: (b) 80

Explanation:
1. Atomic mass of X = 2 \(\times\) 12 = 24.
2. Molar mass of Y = 24 \(\times\) 10/3 = 80.

Teacher's Note:
1. Atomic mass in u equals molar mass in g/mol.
2. Use the ratio carefully.

Question 5: The charge on 1 gram ion of Al\(^{3+}\) is (N\(_A\) = Avogadro number, e = charge on one electron):
(a) (1/3) N\(_A\) e coulomb
(b) (1/9) N\(_A\) e coulomb
(c) 3 N\(_A\) e coulomb
(d) (1/27) N\(_A\) e coulomb
Show Answer & Explanation

Answer: (c) 3 N\(_A\) e coulomb

Explanation:
1. 1 gram ion = 1 mole of ions = N\(_A\) ions.
2. Each ion carries charge 3e, so the total charge is 3 N\(_A\) e.

Teacher's Note:
1. The charge per ion is +3e.
2. Charge = number of ions \(\times\) charge per ion.

Question 6: Suppose the mass of a neutron is taken as half of its actual value and that of a proton as twice its actual value (electron mass negligible). Then the atomic mass of \(^{14}_{6}\)C will be:
(a) 14.28% more
(b) same
(c) 114.28% less
(d) 28.56% less
Show Answer & Explanation

Answer: (a) 14.28% more

Explanation:
1. Original mass number 14 (6 protons and 8 neutrons).
2. New mass = 6 \(\times\) 2 + 8 \(\times\) 0.5 = 16.
3. Change = (16 \(-\) 14)/14 = 14.28% more.

Teacher's Note:
1. Percentage change = change / original \(\times\) 100.
2. Electron mass is negligible.

Question 7: The isotopic abundance of C-12 and C-14 is 98% and 2% by mass respectively. The number of C-14 atoms in a 12 g carbon sample is:
(a) \( 1.032 \times 10^{22} \)
(b) \( 6.02 \times 10^{23} \)
(c) \( 3.01 \times 10^{23} \)
(d) \( 5.88 \times 10^{23} \)
Show Answer & Explanation

Answer: (a) \( 1.032 \times 10^{22} \)

Explanation:
1. Mass of C-14 = 0.02 \(\times\) 12 = 0.24 g.
2. Moles = 0.24/14 = 0.01714.
3. Atoms = 0.01714 \(\times 6.022 \times 10^{23} = 1.032 \times 10^{22}\).

Teacher's Note:
1. The percentages are by mass, not by number.
2. Use the mass number 14 for C-14.

Question 8: Indium (atomic weight 114.8) has two naturally occurring isotopes. The predominant one has isotopic weight 115 and abundance 95.00%. Which isotopic weight is most likely for the other isotope?
(a) 112
(b) 113
(c) 114
(d) 111
Show Answer & Explanation

Answer: (d) 111

Explanation:
1. 114.8 = 0.95 \(\times\) 115 + 0.05 \(\times\) x.
2. 0.05x = 114.8 \(-\) 109.25 = 5.55, so x = 111.

Teacher's Note:
1. Use the weighted average formula.
2. Solve for the unknown isotope mass.

Question 9: The number of molecules of CO\(_2\) present in 44 g of CO\(_2\) is:
(a) \( 3 \times 10^{10} \)
(b) \( 6.0 \times 10^{23} \)
(c) \( 12 \times 10^{23} \)
(d) \( 3 \times 10^{23} \)
Show Answer & Explanation

Answer: (b) \( 6.0 \times 10^{23} \)

Explanation:
1. 44 g is one mole of CO\(_2\), which has N\(_A\) molecules.

Teacher's Note:
1. One mole = N\(_A\) molecules.
2. 44 g/mol is the molar mass of CO\(_2\).

Question 10: The number of moles of ammonia in 4.25 g of ammonia is:
(a) 0.2125
(b) 0.25
(c) 0.236
(d) 0.425
Show Answer & Explanation

Answer: (b) 0.25

Explanation:
1. Moles = 4.25/17 = 0.25.

Teacher's Note:
1. Molar mass of NH\(_3\) = 17 g/mol.
2. Moles = mass / molar mass.

Question 11: Which pair of gases contains the same number of molecules?
(a) 16 g of O\(_2\) and 14 g of N\(_2\)
(b) 8 g of O\(_2\) and 22 g of CO\(_2\)
(c) 28 g of N\(_2\) and 22 g of CO\(_2\)
(d) 32 g of O\(_2\) and 32 g of N\(_2\)
Show Answer & Explanation

Answer: (a) 16 g of O\(_2\) and 14 g of N\(_2\)

Explanation:
1. 16 g O\(_2\) = 0.5 mol and 14 g N\(_2\) = 0.5 mol.
2. The other pairs have unequal moles: 0.25 and 0.5; 1 and 0.5; 1 and 1.14.

Teacher's Note:
1. Same moles means same number of molecules.
2. Convert each mass to moles.

Question 12: The number of electrons in 1.8 mL of H\(_2\)O(l) is about:
(a) \( 6.02 \times 10^{23} \)
(b) \( 3.011 \times 10^{23} \)
(c) \( 60.22 \times 10^{20} \)
(d) \( 0.6022 \times 10^{21} \)
Show Answer & Explanation

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

Explanation:
1. 1.8 mL of water = 1.8 g = 0.1 mol.
2. Electrons per molecule = 10, so total = 1 mol of electrons = \( 6.02 \times 10^{23} \).

Teacher's Note:
1. Density of water is 1 g/mL.
2. Water has 10 electrons per molecule.

Question 13: A sample of ammonium phosphate, (NH\(_4\))\(_3\)PO\(_4\), contains 3.18 mole of H atoms. The number of moles of O atoms in the sample is:
(a) 1.06
(b) 0.265
(c) 3.18
(d) 0.795
Show Answer & Explanation

Answer: (a) 1.06

Explanation:
1. Each formula unit has 12 H and 4 O atoms.
2. Moles of formula units = 3.18/12 = 0.265.
3. Moles of O = 0.265 \(\times\) 4 = 1.06.

Teacher's Note:
1. Use the ratio H : O = 12 : 4 = 3 : 1.
2. So O = 3.18/3 = 1.06.

Question 14: Torr is a unit of:
(a) volume
(b) temperature
(c) density
(d) pressure
Show Answer & Explanation

Answer: (d) pressure

Explanation:
1. 1 torr = 1 mm Hg = 133.3 Pa, a unit of pressure.

Teacher's Note:
1. Named after Torricelli.
2. 1 atm = 760 torr.

Question 15: The atmospheric pressure on Mars is 0.61 kPa. What is this pressure in mm Hg?
(a) 3.2
(b) 0.63
(c) 6.3
(d) 4.6
Show Answer & Explanation

Answer: (d) 4.6

Explanation:
1. 0.61 kPa = 610 Pa.
2. 610/133.3 = 4.6 mm Hg.

Teacher's Note:
1. 1 mm Hg = 133.3 Pa.
2. Remember to convert kPa to Pa first.

Question 16: Which relation connects the Celsius and Fahrenheit scales?
(a) C/5 = (F + 32)/9
(b) C/5 = (F \(-\) 32)/9
(c) C/8 = (F \(-\) 32)/5
(d) C/9 = (F \(-\) 32)/5
Show Answer & Explanation

Answer: (b) C/5 = (F \(-\) 32)/9

Explanation:
1. Freezing and boiling points are 0 and 100 on Celsius and 32 and 212 on Fahrenheit.
2. So 100 Celsius degrees equal 180 Fahrenheit degrees: C/5 = (F \(-\) 32)/9.

Teacher's Note:
1. Check with water: C = 100 gives F = 212.
2. Sizes of degrees: 5 C = 9 F.

Question 17: At what temperature do the Celsius and Fahrenheit scales read the same value?
(a) 60\(^\circ\)
(b) 100\(^\circ\)
(c) 130\(^\circ\)
(d) \(-\)40\(^\circ\)
Show Answer & Explanation

Answer: (d) \(-\)40\(^\circ\)

Explanation:
1. Put C = F = x in C/5 = (F \(-\) 32)/9: 9x = 5x \(-\) 160, so x = \(-\)40.

Teacher's Note:
1. Both scales read \(-\)40.
2. Check: \(-\)40 \(^\circ\)C = \(-\)40 \(^\circ\)F.

Question 18: The value of the universal gas constant R depends on:
(a) volume of the gas
(b) number of moles of the gas
(c) temperature of the gas
(d) units of volume and pressure
Show Answer & Explanation

Answer: (d) units of volume and pressure

Explanation:
1. R is a constant.
2. Its numerical value changes only with the units used (for example 0.0821 L atm or 8.314 J).

Teacher's Note:
1. R = PV/nT has the same physical value for all gases.
2. Only the units change the number.

Question 19: The value of the gas constant in calorie per kelvin per mole is approximately:
(a) 3 cal
(b) 2 cal
(c) 4 cal
(d) 1 cal
Show Answer & Explanation

Answer: (b) 2 cal

Explanation:
1. R = 8.314 J/(K mol) = 8.314/4.184 = 1.987 cal/(K mol), about 2 cal.

Teacher's Note:
1. 1 cal = 4.184 J.
2. R is about 2 cal K\(^{-1}\) mol\(^{-1}\).

Question 20: The value of R in SI units is:
(a) \( 8.314 \times 10^{-7} \) erg K\(^{-1}\) mol\(^{-1}\)
(b) 2 cal K\(^{-1}\) mol\(^{-1}\)
(c) 0.082 L atm K\(^{-1}\) mol\(^{-1}\)
(d) 8.314 J K\(^{-1}\) mol\(^{-1}\)
Show Answer & Explanation

Answer: (d) 8.314 J K\(^{-1}\) mol\(^{-1}\)

Explanation:
1. The SI unit of energy is the joule.
2. So R = 8.314 J K\(^{-1}\) mol\(^{-1}\). In erg it would be \( 8.314 \times 10^{7} \).

Teacher's Note:
1. 1 J = \( 10^7 \) erg.
2. Other values are in non-SI units.

Question 21: The pressure of a gas containing 2 mole in a 44.8 L vessel at 546 K is:
(a) 2 atm
(b) 1 atm
(c) 4 atm
(d) 3 atm
Show Answer & Explanation

Answer: (a) 2 atm

Explanation:
1. P = nRT/V = 2 \(\times\) 0.0821 \(\times\) 546/44.8 = 2.0 atm.

Teacher's Note:
1. 546 K is twice 273 K.
2. Use R = 0.0821 L atm K\(^{-1}\) mol\(^{-1}\).

Question 22: According to the ideal gas laws, the molar volume of a gas is given by:
(a) RT/P
(b) 8RT/PV
(c) RT/PV
(d) 22.4 litre
Show Answer & Explanation

Answer: (a) RT/P

Explanation:
1. For one mole PV = RT, so molar volume V = RT/P.
2. 22.4 L is its value only at STP.

Teacher's Note:
1. Molar volume depends on T and P.
2. 22.4 L holds at 0 \(^\circ\)C and 1 atm.

Question 23: Equal volumes of oxygen and a second gas weigh 1.00 g and 19/8 g respectively under the same conditions. The unknown gas is:
(a) SO\(_2\)
(b) NO
(c) CS\(_2\)
(d) CO
Show Answer & Explanation

Answer: (c) CS\(_2\)

Explanation:
1. Equal volumes mean equal moles, so molar mass ratio = mass ratio.
2. M = 32 \(\times\) (19/8) = 76 g/mol, which is CS\(_2\) (12 + 64).

Teacher's Note:
1. Avogadro's law: equal volumes, equal moles.
2. SO\(_2\) has molar mass 64.

Question 24: Four 1-litre flasks are filled with H\(_2\), He, O\(_2\) and O\(_3\) at the same temperature and pressure. The ratio of the total number of atoms in the flasks is:
(a) 1 : 2 : 2 : 3
(b) 3 : 2 : 2 : 1
(c) 1 : 1 : 1 : 1
(d) 2 : 1 : 2 : 3
Show Answer & Explanation

Answer: (d) 2 : 1 : 2 : 3

Explanation:
1. Equal volumes have equal numbers of molecules.
2. Atoms per molecule: H\(_2\) 2, He 1, O\(_2\) 2, O\(_3\) 3.
3. So the ratio is 2 : 1 : 2 : 3.

Teacher's Note:
1. Use atomicity.
2. He is monatomic.

Question 25: 16 g of an ideal gas SO\(_x\) occupies 5.6 L at STP. The value of x is:
(a) 3
(b) 2
(c) 4
(d) none of these
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. Moles = 5.6/22.4 = 0.25.
2. Molar mass = 16/0.25 = 64 g/mol = 32 + 16x, so x = 2.

Teacher's Note:
1. Molar mass = mass / moles.
2. Sulphur has atomic mass 32.

Question 26: The ratio of the weight of one litre of a gas to the weight of 1.0 L of oxygen, both at STP, is 2.22. The molecular weight of the gas is:
(a) 71.04
(b) 14.002
(c) 35.52
(d) 55.56
Show Answer & Explanation

Answer: (a) 71.04

Explanation:
1. At the same conditions mass ratio = molar mass ratio.
2. M = 2.22 \(\times\) 32 = 71.04 g/mol (Cl\(_2\)).

Teacher's Note:
1. Equal volumes contain equal moles.
2. The gas is chlorine.

Question 27: The approximate weight of \( 1 \times 10^{22} \) molecules of CuSO\(_4\)\(\cdot\)5H\(_2\)O is:
(a) 41.59 g
(b) 415.9 g
(c) 4.159 g
(d) none of these
Show Answer & Explanation

Answer: (c) 4.159 g

Explanation:
1. Molar mass = 63.5 + 32 + 64 + 90 = 249.5 g/mol.
2. Moles = \( 10^{22}/6.022 \times 10^{23} = 0.0166 \).
3. Mass = 0.0166 \(\times\) 249.5 = 4.14 g, closest to 4.159 g.

Teacher's Note:
1. Include water of crystallisation in the molar mass.
2. Estimate to choose the closest option.

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