# AP 9th Class Physical Science Chapter 5 Solution – Atoms and Molecules

## AP 9th Class Physical Science Chapter 5 Solution – Atoms and Molecules

Andhra Pradesh SCERT 9th Class Physical Science Chapter 5 Atoms and Molecules Solution for AP 9th Class Physics/Chemistry Exam. Lots of Students of Andhra Pradesh Board will search on internet for Andhra Pradesh Class 9 Physical Science Textbook Solution or Study Material for AP 9th exam. Here you search will end! Here in this page we have provided for all question answer for AP SCERT Chapter 5 Atoms and Molecules.

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### AP 9th Class Chemistry Chapter 5 Atoms and Molecules Solution

1.) Draw the diagram to show the experimental setup for the law of conservation of mass. (AS5)

Ans: –After taking some amount of copper sulphate in test tube and NaOH solution in flask. Then mixed these two solution, we will see that there is no change in their mass to the previous. So it follows the conservation of mass principal. 2) Explain the process and precautions in verifying law of conservation of mass. (AS1)

Ans: –According to the conservation of mass we know that matter is neither be destroyed or created in any chemical reaction. So it clearly tell us that in a chemical reaction the mass of the reactant always be equal to mass of the products.

3.) 15.9g. of copper sulphate and 10.6g of sodium carbonate react together to give14.2g of sodium sulphate and 12.3g of copper carbonate. Which law of chemical combination is obeyed? How? (AS1 , AS2)

Ans: – As we see that in these reaction the mass of reactants (15.9 +10.6 = 26.5) and the mass of products (14.2 + 12.3 = 26.5) are same so we can say that laws of conservation of mass obeyed here.

4.) Carbon dioxide is added to 112g of calcium oxide. The product formed is 200g of calcium carbonate. Calculate the mass carbon dioxide used. Which law of chemical combination will govern your answer.(AS1 , AS2)

Ans: –  CO2(x) + CaO (112g) —> CaCO3 (200g).

According to law of conservation of mass,x +112 = 200; or x = 108g.

So law of conservation of mass govern my answer.

5.) 0.24g sample of compound of oxygen and boron was found by analysis to contain 0.144g of oxygen and 0.096g of boron. Calculate the percentage composition of the compound by weight. (AS1)

Ans: –As we all know that % of composition of any compound = (mass of the compound / mass of the volume) ×100.

So % of composition of oxygen = 0.144/0.24 × 100 = 60%.

And the % composition of boron = 0.096/0.24 × 100 = 40%.

6.) In a class, a teacher asked to write the molecular formula of oxygen Samhita wrote the formula as O2 and Priyanka as O. which one is correct? State the reason.(AS1,AS2)

Ans: –  A molecule present in nature are diatomic, so as Samhita wrote O2 which clarify the actual representation of oxygen for this reason it will be correct.

7.) Imagine what would happen if we do not have standard symbols for elements?(AS2)

Ans:-  If there is no any certain symbols for an element it will create problem to calculate its chemical property as without knowing the proper symbol it hamper the chemical reaction.

8.) Mohith said “H2 differs from 2H”. Justify. (AS1)

Ans: –H2 tells us that hydrogen is bonded with two atom whereas 2H means two atom of hydrogen which are not bonded.

9.) Lakshmi gives a statement “CO and Co both represents element”. Is it correct? State reason.(AS1,AS2)

Ans: – CO and Co doesn’t represent the same as Co is symbol of element but CO which is acompound. In these case the second letter defined that what will be its character.

10) The formula of water molecule is H2O. What information you get from this formula.(AS1)

Ans: – The formula of water molecules H2O tells us that there is two hydrogen atom and one oxygen atom in the molecule.  Making bond to each other they form water molecule.

11) How would you write 2 molecules of oxygen and 5 molecules of Nitrogen.(AS1)

Ans: – The 2 molecules of oxygen is represented as O2.

And 5 molecules of Nitrogen is represented as N5.

12) The formula of a metal oxide is MO. Then write the formula of its chloride.(AS1)

Ans: – As the formula of metal oxide is MO so its chloride formula will be MOCl2.

13) Formula of calcium hydroxide is Ca (OH)2and zinc phosphate is Zn3(PO4)2. Then write the formula to calcium phosphate.(AS1)

Ans: – As the valency of Ca is 2 and PO4 is 3 so the formula of calcium phosphate is Ca3 (PO4)2.

14.) Find out the chemical names and formulae for the following common householdsubstances. (AS1)

a) Common salt b) baking soda c) washing soda d) vinegar

Ans: –a) The chemical name and formula of common salt is Sodium Chloride and NaCl.

b) The chemical name of baking soda and its formula is sodium bicarbonate and NaHCO3.

c) Washing soda —> Sodium carbonate, Na2CO3.

d) The chemical name of vinegar is acetic acid and formula is CH3COOH.

15.) Calculate the mass of the following (AS1)

a) 0.5 mole of N2 gas. b) 0.5 mole of N atoms.

c) 3.011 X 1023 number of N atoms. d) 6.022 X 1023 number of N2 molecules.

Ans: –a) We know that,

no moles ofN2 = mass of N2 / molar mass of N2

Or, mass of N2= molar mass of N2 × no of moles of N2 = 2 ×14 ×0.5 (as nitrogen is diatomic so molar mass 2×14)

Or, mass of N2 is 14g.

b) Similarly the mass of N is (0.5 × 14=7) 7g.

c) As we know that 1 mole is equal to 6.023×10^23 no of atoms so 3.011×10^23 mean 0.5 mol.

So mass of N atoms is (0.5 × 14= 7) 7g.

d) From the mole concept we know that in 1 mole there are 6.23 × 10^23 no of atoms.

So mass of N2 is (2×14×1=28) 28g.

16.) Calculate the number of particles in each of the following (AS1)

a) 46g of Na b) 8g of O2

c) 0.1 mole of hydrogen.

Ans: –a) In one mole of Na there are 6.023 × 10^23 no of particle.

Here 46g Na means 2mole of Na as atomic mass of Na is 23. So the no of particle will be 2×6.023×10^23 or 12.046×10^23.

b) In one mole or 32g of oxygen there are 6.023×10^23 no of particle.

So the no particle in 8g O2 is (8/32×6.023×10^23)= 1.5 ×10^23.

c)In one mole of hydrogen atom there are 6.023 ×10^23 no of particle present.

Then no of particle in 0.1 mole is (0.1×6.023×10^23) = 6.023×10^24.

17.) Convert into mole (AS1)

a) 12g of O2 gas. b) 20g of water. c) 22g of carbon dioxide.

Ans: – a) One mole of oxygen = 32g of oxygen (a.w of O2 is 32).

So 12g of O2 means 12/32 or 0.37 mole.

b)Atomic weight of H2O is 18 or 18g H2O means one mole.

So 20g of H2O is equal to (20/18) 1.11mole.

c) 44g of CO2 = 1 mole. (a.w of CO2 is 44).

So 22g of CO2 is (22/44) 0.5 mole.

18.) Write the valences of Fe in FeCl2and FeCl3. (AS1)

Ans: – The valences of Fe in FeCl2 and FeCl3 is +2 and +3.

19.) Calculate the molar mass of Sulphuric acid (H2SO4) and glucose (C6H12O6) (AS1)

Ans: – Molar mass of H2SO4 = 2×1 + 32 + 4×16 = 98u.

Molar mass of C6H12O6 = 6×12 + 12×1 + 6×16 = 108u.

20.) Which has more number of atoms – 100g of sodium or 100g of iron? Justify youranswer. (atomic mass of sodium = 23u, atomic mass of iron = 56u) (AS1)

Ans: –No of atoms in 100g of sodium is 100/23×6.023×10^23 = 2.613 × 10^24.

And no of atoms in 100g of iron is 100/56 × 6.023×10^23 = 1.07 ×10^24.

From the upper calculation we came to this conclusion that 100g sodium has more atoms.

21.) Complete the following table. (AS1)

Ans:-

 Anions Cations Chloride Hydroxide Nitrate Sulphate Carbonate Phosphate Sodium NaCl NaOH NaNO3 NaSO4 NaCO3 Na3PO4 Magnesium MgCl2 Mg(OH)2 Mg(NO3)2 MgSO4 MgCO3 Mg3(PO4)2 Calcium CaCl2 Ca(OH)2 Ca(NO3)2 CaSO4 CaCO3 Ca3(PO4)2 Aluminium AlCl3 Al(OH)3 Al(NO3)3 Al(SO4)3 Al(CO3)3 AlPO4 Ammonium NH4Cl NH4(OH) NH4NO3 (NH4)2SO4 (NH4)2CO3 (NH4)3PO4

22.) Fill the following table (AS1)

Ans:-

 Sl. No Name Symbol Molar mass No of particles present in molar mass 1 Atomic Oxygen O 16g 6.023×10^23atoms of oxygen 2 Molecular oxygen O2 32g 6.023×10^23 molecules of oxygen 3 Sodium Na 23g 6.023×10^23 unit sodium 4 Sodium ion Na+ 23g 6.023×10^23 units sodium ion 5 Sodium Chloride NaCl 58.5g 6.023×10^23 units of NaCl 6 Water H2O 18g 6.023×10^23 units of water

Updated: November 20, 2021 — 1:44 pm