Class 9 Science Chapter 3 HOTS Questions - Atoms and Molecules

Q1: (a) An element shows variable valencies 4 and 6. Write the formulae of its two oxides.
(b) An element forms an oxide A₂O₅.
(i) What is the valency of the element A ?
(ii) What will be the formula of the chloride of the element ?
Ans: (a) Let the element be represented by the symbol E.
Formula of oxide in which valency of E is 4 = E₂O₄ or EO₂
Formula of oxide in which valency of E is 6 = E₂O₆ or EO₃
(b) Formula of oxide of the element = A₂O₅
(i) The valency of the element A in the oxide = 5⁺
(ii) The formula of the chloride of the element A = ACl₅.  

Q2: (a) Why does not the atomic mass of an element represent the actual mass of its atom ?
(b) The atomic mass of an element is in fraction. What does it mean ?
(c) Why is the value of Avogadro’s number 6.022 x 10²³ and not any other value ?
(d) Does one gram mole of a gas occupy 24.4 L under all conditions of temperature and pressure ?
Ans: (a) Atoms of different elements are very small in size and their actual mass are extremely small. For example, the mass of an atom of hydrogen is 1.67 x 10^-27 kg. To solve this problem, we consider the relative atomic masses of the elements. The relative atomic mas of hydrogen is 1 u and its corresponding gram atomic mass is 1 g.
(b) If the atomic mass of an element is in fraction, this means that it exists in the form of isotopes. The atomic mass is the average atomic mass and is generally fractional.
(c) It represents the number of atoms in one gram atom of an element or the number of molecules in one gram mole of a compound. If we divide the atomic mass of an element by actual mass of its atom, the value is 6.022 x 10²³. Similarly, by dividing the molecular mass of a compound by the actual mass of its molecule, the same result is obtained.
(d) No, one gram mole of a gas occupies a volume of 22.4 L only under N.T.P. conditions i.e. at 273 K temperature and under 760 mm pressure.

Q3: What weight of calcium contains the same number of atoms as are present in 3.2 g of sulphur ?
Ans: Step I. No. of atoms in 3.2 g of sulphur
Gram atomic mass of S = 32 g
32 g of sulphur contain atoms = 6.022 x 10²³ 
3.2 g of sulphur contain atoms

Step II. Weight of 6.022 x 10²² atoms of calcium
Gram atomic mass of Ca = 40 g
6.022 x 10²³ atoms of Ca weigh = 40 g 

6.022 x 10²² atoms of Ca weigh

Q4: In magnesium sulphide, the ratio by mass of Mg and S is 3 : 4. What is the ratio of the number of Mg and S atoms ?
Ans: Formula of magnesium sulphide = MgS
Ratio by mass of Mg and S  = 3.4

∴ Ratio of the number of Mg and S is

Q5: On analysing an impure sample of sodium chloride, the percentage of chlorine was found to be 45.5 What is the percentage of pure sodium chloride in the sample ?
Ans: Molecular (or Formula) mass of pure NaCl = Atomic mass of Na + Atomic mass of CI
= 23 + 35-5 = 58-5 u


Q6: A flask P contains 0.5 g mole of oxygen gas. Another flask Q contains 0.4 mole of ozone gas. Which of the two flasks contains greater number of oxygen atoms ?
Ans:
1 molecule of oxygen (O₂) = 2 atoms of oxygen
1 molecule of ozone (O₃) = 3 atoms of oxygen
In flask P : 1 mole of oxygen gas = 6.022 x 10²³ molecules
0.5 mole of oxygen gas = 6.022 x 10²³ x 0.5 molecules
= 6.022 x 10²³ x 0.5 x 2 atoms = 6.022 x 10²³ atoms
In flask Q : I mole of ozone gas = 6.022 x 10²³ molecules
0.4 mole of ozone gas = 6.022 x 10²³ x 0.4 molecules
= 6.022 x 10²³ x 0.4 x 3 atoms = 7.23 x 10²³ atoms
∴ Flask Q has a greater number of oxygen atoms as compared to flask P.

Q7: Silicon forms a compound with chlorine in which 5.6 g of silicon is combined with 21.3 g of chlorine..
Calculate the formula of the compound (Atomic mass : Si = 28 : Cl = 35-5).
Ans:
The simplest whole number ratios of different elements are : Si : Cl : : 1 : 3
The formula of the compound = SiCl₃.

Q8: A colourless liquid is thought to be a pure compound. Analysis of three samples of the material yield the following results.

Could the material be a pure compound?
Ans: Analysis
Yes, the material is a pure compound as all the three samples have the same composition.

Q9. What is the fraction of the mass of water due to neutrons? 
Ans: Mass of one mole (Avogadro Number) of neutrons ~ 1 g

Q10: Calculate the number of electrons present in 15.4 g of carbon tetrachloride (CCl4).
Ans: Number of moles of CCl₄ =
∵ = 0.1 mole
1 mole of CCl₄ = 6.02 2 × 10²³ molecules of CCl₄
∴ 0.1 mole of CCl₄ = 0.1 × 6.022 × 10²³ molecules of CCl₄
= 6.022 × 10²² molecules of CCl₄
We know that one atom of carbon has 6 electrons and one atom of chlorine has 17 electrons. Therefore, one molecule of CCl₄ will contain 6 + (4 × 17) = 74 electrons.
∴ Number of electrons in 6.022 × 10²² molecules of CCl₄
= 74 × 6.0²² × 10²² electrons
= 445.6 × 10²² electrons
= 4.456 × 10²⁴ electrons

Q11: A big drop of water has volume 1.0 mL. How many molecules of water are there in this drop, If the density of water is lg/mL?
Ans: Volume of drop of water = 1.0 mL
Density of water = 1.0 g/mL
∴ Mass of drop of water = Volume × Density = 1.0 g
Molecular mass of H₂O = 2 × lu + 1 × 16u = 18u
Gram molecular mass of water =18 g/mol
18 g of water contains = 6.02 2 × 10²³ molecules
∴ 1 g of water contains  molecules
= 3.34 × 10²² molecules

Q12: You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting?
Ans: On heating the powder, it will char if it is a sugar.
Alternatively, the powder may be dissolved in water and checked for its conduction of electricity. If it conductsm it is salt.