CAN ANYONE GIVE ME ATLEAST 25 QUESTIONS BASED ON MOLE CONCEPT WITH ANSWERS. PLEASE ITS VERY URGENT.
Answers
(a) 6.022 × 1023 atoms
(b) one atom
(c) 35.5 g of Cl
(d) All of the above.
Answer: Both (a) and (b) are correct. 1 mol of Cl atom = 36.5 g of Cl (molar mass)
= 6.02 × 1023 atoms.
Q4: 1.0 mole of Chlorine molecule (Cl2) contains
(i) how many number of molecules.
(ii) how many number of atoms.
(iii) how much it weighs.
Answer: (i) 1.0 mol of Cl2 contains 6.022 × 1023molecules.
(ii) One molecule of Cl2 contains 2 toms of Cl. ∴ Cl2 contains 2 × 6.022 × 1023 atoms
i.e. 12.44 × 1023 atoms
(iii) Molar mass of Cl is 35.5 gm/mol. 1.0 mol of Cl2 weighs = 2 × 35.5 = 71.1 g
Mole in terms of number,
1 Mole of particle = 6.022 × 1023particles
Q5: Which of the following is correct option?
1.0 mole of NH3 (ammonia) contains ...
(a) 6.022 × 1023 molecules
(b) 4 mol of atoms
(c) 1 mol of Nitrogen atoms
(d) 3 × 6.022 × 1023 of H atoms
Answer: All of the above options are correct.
Q6: What is atomic mass unit (amu)?
Answer: Atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon-12 atom. One atomic mass unit also called one Dalton.
Mass of one mole of C-12 atoms = 12 g = mass of 6.022 × 1023 C-12 atoms
1 amu = 1g per mol = 1/NA = 1/ (6.022 × 1023) = 1.66 × 10-24g.
Q7: What is the mass of one 12C atom? Express it in grams as well as in amu.
Answer: one mole of C-12 atoms = 12 g = mass of 6.022 × 1023 C-12 atoms
∴ mass of 1 C-12 atom = 12g ÷ 6.022 × 1023 = 1.994 × 10-23g
or = 12g ÷ 6.022 × 1023 = 12g × 1.66 × 10-24g = 12 amu.
Q8: What is molar mass?
Answer: The mass of one mole of an element or one mole of compound is referred as molar mass.
It is expressed as g mol-1.
Example:
molar mass of Mg = 24 g mol-1.
molar mass of methane (CH4) = (12 + 4) g mol-1 = 16 g mol-1.
Q9: What is Gram atomic mass or molar mass of an element?
Answer: Gram atomic mass or molar mass of an element is mass of 1 mol of atoms or atomic mass expressed in grams. For example, atomic mass of Mg = 24u, therefore, molar mass of Mg is 24 grams per mol. Molar mass of an element is also called one gram atom.
Q10: What is Gram molecular mass or molar mass of molecular substance?
Answer: Gram molecular mass or molar mass of a molecular substance is the mass of 1 mol of molecules or molecular mass expressed is grams. For example, molecular mass of H2O is 18u (2u + 16u), therefore, molar mass of H2O is 18 g mol-1.
Mole in terms of mass,
No. of Moles = Mass ÷ Molar mass
Q11: In 14g of N2(nitrogen gas), calculate
(i) number of moles (take molar mass of 28 g mol-1)
(ii) number of molecules
(iii) number of atoms
Answer:
(i) No. of moles = 14g ÷ 28 g mol-1 = 0.5 moles
(ii) No. of molecules = (number of moles) × NA= 0.5 × 6.022 × 1023 = 3.011 × 1023 molecules
(iii) One molecule of N2 contains 2 N atoms.
No. of atoms = 2 × 3.011 × 1023 = 6.022 × 1023atoms = 1 NA atoms.
Question 1.
Calculate the mass of 6.022 × 1023 molecule of Calcium carbonate (CaCO3).
Solution —
Molar mass (Molecular mass in gram) of CaCO3 = 40+12+3×16 = 100 g
No. of moles of CaCO3
= No. of molecules/Avogadro constant
= 6.022 × 1023/ 6.022 × 1023
= 1 mole
Mass of CaCO3
= No. of moles × molar mass
= 1 × 100 g = 100 g.
Question 2.
Calculate the mass of 12.044 × 1023 carbon atoms.
Solution —
No. of moles of Carbon atoms
= No. of atoms/Avogadro constant
= 12.044 × 1023/6.022 × 1023
= 2 mole
Mass of carbon atoms
= No. of moles × atomic mass
= 2 × 12
= 24 g.
Question 3.
Calculate the number of oxygen atoms in 1 mole of O2.
Solution —
1 molecule of O2 = 2 oxygen atoms
So, 1 mole of O2 = 2 mole oxygen atoms
= 2 × 6.022 × 1023 = 12.044 ×1023 oxygen atoms.
Question 4.
Calculate the number of Cu atoms in 0.635g of Cu.
Solution —
No. of moles of Cu
= Mass of Cu/Atomic mass
= 0.635/63.5
=0.01 mole
No. of Cu atoms
= No. of moles × Avogadro constant
= 0.01 × 6.022 × 1023
= 6.022 × 1023 Cu atoms.
Question 5.
Calculate the number of molecules in 11.2 liters of SO2 gas at NTP.
Solution —
1 mole of SO2 = 22.4 L (at NTP)
=> 11.4 L of SO2 = 0.5 mole SO2 = 0.5× 6.022×1023 = 3.011×1023 SO2 molecules.
Question 6.
An atom of some element X weighs 6.644 × 10-23 g. Calculate the number of gram-atoms in 40 kg of it.
Solution —
Mass of 1 mole X atoms
= mass of 1 atom × Avogadro constant
= 6.644 × 10-23 × 6.022×1023
= 40 g
So, the atomic mass of X = 40
No. of gram-atoms (or moles) of X
= mass of X / atomic mass
= 40 × 1000/40 = 1000.
Question 7.
An atom of some element X weighs 6.644 × 10-23 g. Calculate the number of gram-atoms in 40 kg of it.
Solution —
Molecular mass of CO2 = 12 + 2 × 16 = 44
Total no. of moles in 200mg CO2
= Mass of CO2/Molecular mass
= 200 × 10-3 g/44
= 0.00454
No. of moles removed
= 1021/6.022×1023
= 0.00166
No. of moles of CO2 left
= 0.00454 – 0.00166
= 0.00288.
Question 8.
Calculate the volume occupied by 1 mole atom of (i) monoatomic gas, and (ii) diatomic gas at NTP.
Solution —
1 mole atom of monoatomic gas occupies 22.4 L at NTP, and
1 mole of diatomic gas (1 molecule contains 2 atoms) occupies 11.4 L at NTP.
Question 9.
Calculate the volume of 20g H2 at NTP.
Solution —
No. of moles of H2 = 20/2 =10
Volume of any ideal gas at NTP
= No. of moles × 22.4 L
= 10 × 22.4
= 224 L.
Question 10.
What is the volume occupied by 6.022×1023molecules of any gas at NTP?
Solution —
6.022 × 1023 molecules = 1 mole molecules, and
1 mole molecules of any ideal gas occupies 22.4 L at NTP.
Question 11.
Calculate the number of atoms in 5.6 liters of a (i) monoatomic, and (ii) diatomic gas at NTP.
Solution —
No. of moles in 5.6 L gas at NTP = 5.6/22.4 = 0.25
No. of molecules in 5.6 L gas
= 0.25 × 6.022 × 1023
= 1.5 × 1023 molecules
In monoatomic gases, No. of atoms = No. of molecules = 1.5 × 1023.
In diatomic gases, No. of atoms = 2 × No. of molecules = 2 × 1.5 × 1023 = 3.0× 1023.
Question 12.
Calculate the number of sulphate (SO42-) ions in 100 mL of 0.001 M H2SO4 solution.
Solution —
Molarity = No. of moles of solute/ Volume of solution in Liters
=> No. of moles of solute (H2SO4) = Molarity × volume of solution in Liters
= 0.001 × 0.1 = 0.0001
1 molecule of H2SO4 contains 1 SO42- ion
=> 0.0001 mole of H2SO4 contains 0.0001 mole SO42-
∴ No. of sulphate (SO42-) ions = 0.0001 × 6.022 × 1023 = 6.022 × 1019.
Question 13.
Calculate the number of atoms in 100 u of He.
Solution —
Atomic mass of He = 4 u
So, mass of one He atom = 4 u
∴ No. of atoms in 100 u of He = 100/4 = 25 He atoms.
Question 14.
If a mole were to contain 1× 1024 particles, what would be the mass of (i) one mole of oxygen, and (ii) a single oxygen molecule?
Solution —
Mass of one mole of oxygen molecule (O2)
= molecular mass of oxygen molecule (O2) in gram
= 32 g.
Mass of a single oxygen molecule
= 32/1 × 1024
= 3.2 × 10-23g.
Question 15.
Calculate the standard molar volume of oxygen gas. The density of O2 gas at NTP is 1.429g/L.
Solution —
Standard molar volume
= volume occupied by 1 mole (i.e., 32g) of the O2 gas at NTP
= Mass/density [∵ density = mass/volume => volume = mass/density]
= 32/1.429
= 22.39 Liters.