CBSE Class 9 Science Atoms and Molecules Worksheet Set 08

LONG ANSWER TYPE QUESTIONS
 

Question. Dalton’s atomic theory is contradicted by the formula of sucrose (\(\text{C}_{12}\text{H}_{22}\text{O}_{11}\)). Justify the statement.
Answer: Dalton’s atomic theory states that atoms of different elements combine together in simple whole number ratio. In the formula of \(\text{C}_{12}\text{H}_{22}\text{O}_{11}\) the carbon, hydrogen and oxygen combine in whole number ratio but the ratio is not simple.

Question. How many moles are present in 4 g of sodium hydroxide?
Answer: Gram molar mass of \(\text{NaOH} = 23 + 16 + 1 = 40\text{ g}\)
\(40\text{ g of NaOH} = 1\text{ mol}\)
\(\therefore 1\text{ g of NaOH} = \frac{1}{40}\text{ mol}\)
\(\therefore 4\text{ g of NaOH} = \frac{1}{40} \times 4\text{ mol} = 0.1\text{ mol}\)

Question. State and explain the law of Conservation of mass. Give an example to show that the law applies to physical changes also.
Answer: Law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction. However, this law applies to physical changes also. For example, when ice melts into water, the mass of ice equals to the mass of water, i.e., the mass is conserved. This verifies the law of conservation of mass.

Question. A sample of ammonia weighs 3.00 g. What mass of sulphur trioxide contains the same number of molecules as are in 3.00 g ammonia?
Answer: Number of moles of ammonia in \(3.00\text{ g} = 0.1764\text{ mol}\)
Molecular mass of \(\text{SO}_3 = 1 \times 32\text{ u} + 3 \times 16\text{ u} = 80\text{ u}\)
\(1\text{ mole of SO}_3\text{ weighs } 80\text{ g}\)
\(\therefore 0.1764\text{ moles weigh} = 80 \times 0.1764\text{ g} = 14.11\text{ g}\)

Question. Carbon dioxide produced by action of dilute hydrochloric acid on potassium hydrogen carbonate is moist whereas that produced by heating potassium hydrogen carbonate is dry. What would be the difference in the composition of carbon dioxide in the two cases? State the associated law.
Answer: The composition of \(\text{CO}_2\) in both the cases would be same, i.e., the carbon and oxygen will combine in the same ratio \(1 : 2\).
The law associated is law of constant proportion.

Question. How many atoms would be present in a black dot marked on the paper with graphite pencil as a full stop at the end of a sentence. [Given mass of a dot = \(10^{-18}\text{ g}\)]
Answer: \(1\text{ mole of carbon atoms weigh} = 12\text{ g}\)
Also, \(1\text{ mole of carbon atoms} = 6.022 \times 10^{23}\text{ atoms}\)
Thus, \(12\text{ g of carbon atoms has } 6.022 \times 10^{23}\text{ atoms}\).
\(\therefore 10^{-18}\text{ g of carbon will have } \frac{6.022 \times 10^{23} \times 10^{-18}}{12}\text{ carbon atoms}\)
\(= 5.02 \times 10^{4}\text{ carbon atoms}\).

Question. Calculate the molecular mass of the following:
(a) \(\text{H}_2\text{CO}_3\)
(b) \(\text{C}_2\text{H}_5\text{OH}\)
(c) \(\text{MgSO}_4\)
Answer:
(a) Molecular mass of \(\text{H}_2\text{CO}_3 = 2 \times 1 + 1 \times 12 + 3 \times 16\)
\(= 2 + 12 + 48\)
\(= 62\text{ u}\)
(b) Molecular mass of \(\text{C}_2\text{H}_5\text{OH} = 2 \times 12 + 5 \times 1 + 1 \times 16 + 1\)
\(= 24 + 5 + 16 + 1\)
\(= 46\text{ u}\)
(c) Molecular mass of \(\text{MgSO}_4 = 1 \times 24 + 1 \times 32 + 4 \times 16\)
\(= 24 + 32 + 64\)
\(= 120\text{ u}\)

Question. What are ionic and molecular compounds? Give examples.
Answer: Atoms of different elements join together in definite proportions to form molecules of compounds. For example, water, ammonia, carbon dioxide. Compounds composed of metals and non-metals contain charged species. The charged species are known as ions. An ion is a charged particle and can be negatively or positively charged. A negatively charged ion is called an anion and the positively charged ion is called cation. For example, sodium chloride, calcium oxide.

Question. Give three significances of mole.
Answer:
• One mole represents \(6.022 \times 10^{23}\) entities of a substance.
• One mole of an element contains \(6.022 \times 10^{23}\) atoms of the element.
• One mole of a substance represents one gram formula mass of the substance.

Question. How many molecules are there in 0.5 mol of water?
Answer: \(1\text{ mol of water contains } 6.022 \times 10^{23}\text{ molecules}\)
\(\therefore 0.5\text{ mol of water contains } 6.022 \times 10^{23} \times 0.5\text{ molecules}\)
\(= 3.011 \times 10^{23}\text{ molecules}\)

Question. Calculate the number of moles present in:
(i) \(3.011 \times 10^{23}\) number of oxygen atoms.
(ii) 60 g of calcium
[Given that atomic mass of Ca = 40 u, Avogadro No. = \(6.022 \times 10^{23}\)]
Answer:
(i) \(1\text{ mole of oxygen contains } 6.022 \times 10^{23}\text{ atoms}\)
\(\therefore 6.022 \times 10^{23}\text{ atoms of oxygen} = 1\text{ mol}\)
\(1\text{ atom of oxygen} = \frac{1}{6.022 \times 10^{23}}\text{ mol}\)
\(\therefore 3.011 \times 10^{23}\text{ atoms of oxygen} = \frac{1 \times 3.011 \times 10^{23}}{6.022 \times 10^{23}}\text{ mol} = 0.5\text{ mol}\)
(ii) Atomic mass of Ca = 40 u
\(40\text{ g of calcium} = 1\text{ mol}\)
\(60\text{ g of calcium} = \frac{60}{40}\text{ mol} = 1.5\text{ mol}\)

Question. Calculate the number of particles in each of the following:
(a) 46 g of Na atom
(b) 8 g of \(\text{O}_2\) molecules
(c) 0.1 moles of carbon atom
Answer:
(a) No. of moles of sodium = \(\frac{46}{23} = 2\text{ moles}\)
We know that one mole of sodium contains \(6.022 \times 10^{23}\text{ atoms}\).
\(\therefore 2\text{ moles of sodium contain} = 2 \times 6.022 \times 10^{23}\text{ atoms} = 1.204 \times 10^{24}\text{ atoms}\)
(b) \(1\text{ mole of oxygen} = 32\text{ g}\)
\(32\text{ g of }\text{O}_2\text{ contains } 6.022 \times 10^{23}\text{ molecules}\)
\(\therefore 8\text{ g of }\text{O}_2\text{ contains} = \frac{6.022 \times 10^{23} \times 8}{32}\text{ molecules} = 1.51 \times 10^{23}\text{ molecules}\)
(c) \(1\text{ mole of carbon atoms contains } 6.022 \times 10^{23}\text{ atoms}\)
\(\dots 0.1\text{ mole of carbon atoms contains} = 6.022 \times 10^{23} \times 0.1\text{ atoms} = 6.022 \times 10^{22}\text{ atoms}\)

Question. State and explain the law of Constant composition or Law of definite proportion
Answer: This law deals with the composition of elements present in a given compound. It was put forward by J.L Proust . It states that 'a chemical compound is always made up of the same elements combined together in the same fixed proportion by mass'
For example water obtained from many sources like rain, river, tap etc is always made up of the same elements hydrogen and oxygen combined together in the same fixed proportion of \(2 : 16\) i.e., \(1 : 8\) by mass

Question. Find the ratio of mass of the combining elements in the following compounds:
(a) \(\text{CaCO}_3\)
(b) \(\text{H}_2\text{SO}_4\)
Answer:
(a) \(\text{CaCO}_3\)
\(\text{Ca} : \text{C} : \text{O} \times 3\)
\(40 : 12 : 16 \times 3\)
\(40 : 12 : 48\)
\(10 : 3 : 12\)
(b) \(\text{H}_2\text{SO}_4\)
\(\text{H} \times 2 : \text{S} : \text{O} \times 4\)
\(1 \times 2 : 32 : 16 \times 4\)
\(2 : 32 : 64\)
\(1 : 16 : 32\)

Question. Verify by calculating that
(a) 5 moles of \(\text{CO}_2\) and 5 moles of \(\text{H}_2\text{O}\) do not have the same mass.
(b) 240 g of calcium and 240 g of magnesium elements have a mole ratio of \(3 : 5\).
Answer:
(a) \(\text{CO}_2\text{ has molar mass} = 44\text{ g mol}^{-1}\)
\(5\text{ moles of }\text{CO}_2\text{ have mass} = 44 \times 5 = 220\text{ g}\)
\(\text{H}_2\text{O}\text{ has molar mass} = 18\text{ g mol}^{-1}\)
\(5\text{ moles of }\text{H}_2\text{O}\text{ have mass} = 18 \times 5\text{ g} = 90\text{ g}\)
(b) Number of moles in \(240\text{ g Ca metal} = \frac{240}{40} = 6\)
Number of moles in \(240\text{ g of Mg metal} = \frac{240}{24} = 10\)
Ratio is \(6 : 10\)
or, \(3 : 5\)

Question. Calculate the ratio between the mass of one atom of hydrogen and mass of one atom of silver.
Answer:
\(1\text{ mole of H atoms} = 1\text{ g}\)
\(1\text{ mole of H atoms} = 6.022 \times 10^{23}\text{ atoms}\).
Mass of \(6.022 \times 10^{23}\text{ atoms of H} = 1\text{ g}\)
Mass of one atom of H \(= \frac{1}{6.022 \times 10^{23}} = 1.66 \times 10^{-24}\text{ g}\)
\(1\text{ mole silver atoms contain } 6.022 \times 10^{23}\text{ atoms}\)
Mass of one silver atom \(= \frac{108}{6.022 \times 10^{23}}\text{ g} = 1.793 \times 10^{-22}\text{ g}\)
Ratio between masses of Ag and H \(= \frac{1.793 \times 10^{-22}}{1.66 \times 10^{-24}} = 108\)

Question. Which of the following would weigh the highest: 2 mole \(\text{CO}_2\), 2 mole \(\text{CaCO}_3\), 10 mole \(\text{H}_2\text{O}\)?
Answer:
\(2\text{ mole }\text{CO}_2\text{ weighs } 2 \times 44\text{ g} = 88\text{ g}\)
\(2\text{ mole }\text{CaCO}_3\text{ weighs } 2 \times 100\text{ g} = 200\text{ g}\)
\(10\text{ mole }\text{H}_2\text{O}\text{ weighs } 10 \times 18\text{ g} = 180\text{ g}\)
Hence 2 mole \(\text{CaCO}_3\) weighs most

Question. Write molecular formulae for
(i) Copper (II) nitrate      (ii) Aluminium (III) nitrate      (iii) Iron (III) sulphide
Answer:
(i) \(\begin{array}{cc} \text{Cu} & \text{NO}_3 \\ 2+ & 1- \end{array} \implies \text{Cu(NO}_3)_2\)
(ii) \(\begin{array}{cc} \text{Al} & \text{NO}_3 \\ 3+ & 1- \end{array} \implies \text{Al(NO}_3)_3\)
(iii) \(\begin{array}{cc} \text{Fe} & \text{S} \\ 3+ & 2- \end{array} \implies \text{Fe}_2\text{S}_3\)

Question. Compute the ratio by mass of the combining elements in the following:
(i) Ammonia
(ii) Magnesium sulphide
Answer:
(i) \(\text{N} : \text{H} = 14 : 3\)
(ii) \(\text{Mg} : \text{S} = 24 : 32\text{ or } 3 : 4\)

Question. Write the formula of the compounds formed by the sets of elements
(i) Calcium and fluorine
(ii) Hydrogen and sulphur
(iii) Sodium and oxygen
Answer:
(i) \(\begin{array}{cc} \text{Ca} & \text{F} \\ 2+ & 1- \end{array} \implies \text{CaF}_2\)
(ii) \(\begin{array}{cc} \text{H} & \text{S} \\ 1+ & 2- \end{array} \implies \text{H}_2\text{S}\)
(iii) \(\begin{array}{cc} \text{Na} & \text{O} \\ 1+ & 2- \end{array} \implies \text{Na}_2\text{O}\)

Question. Fill the missing data in the following table:

Answer:

Question. Two containers A and B are filled with 5 moles of carbon atoms and 5 moles of sodium atoms respectively. Which one is heavier?
Which one contains more number of atoms?
Answer:
Container A: \(5\text{ moles carbon} = 5 \times 12 = 60\text{ g}\)
Container B: \(5\text{ moles sodium} = 5 \times 23 = 115\text{ g}\)
Container B is heavier.
Both contain the same number of atoms \(= 5 \times 6.022 \times 10^{23}\)

Question. Fill in the blanks:
1. In a chemical reaction, the sum of masses of reactants and products remain unchanged. This is called…………………….
2. A group of atoms carrying a fixed charge on them is called……………….
3. Formula of sodium carbonate is…………………
4. Formula of common salt is………………
5. Symbol for mercury is ……………..
Answer:
1. Law of conservation of mass
2. Polyatomic ion
3. \(\text{Na}_2\text{CO}_3\)
4. \(\text{NaCl}\)
5. \(\text{Hg}\)

Question. Find out the number of atoms present in (i) Phosphate ion (ii) Sulphuric acid
Answer:
(i) Phosphate ion is \(\text{PO}_4^{3-}\), number of atoms is 5
(ii) Sulphuric acid is \(\text{H}_2\text{SO}_4\), number of atoms is 7

Question. \(10^{22}\) atoms of an element X are found to have a mass of 930 mg. Calculate the molar mass of the element X.
Answer:
\(10^{22}\text{ atoms of the element have mass } 930\text{ mg} = 0.930\text{ g}\)
Molar mass i.e., \(6.022 \times 10^{23}\text{ atoms of the element will have mass}\)
\(= \frac{0.930 \times 6.022 \times 10^{23}}{10^{22}} = 56\text{ g}\)