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Science Answer the questions.[Che-3M]

Atoms and Molecules · CBSE STD 9 (CBSE - English Medium). 15 questions.

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Answer the questions.[Che-3M]

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15 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
When 5g of calcium is burnt in 2g of oxygen, then 7g of calcium oxide is produced. What mass of calcium oxide will be produced when 5g of calcium is burnt in 20g of oxygen? Which law of chemical combination will govern your answer?
Answer
When 5gm of calcium is burnt in 2gm of oxygen, then 7gm of calcium oxide is formed. So, calcium and oxygen combine in the fixed proportion of 5 : 2 by mass. Now, when 5gm of calcium is burnt in 20gm of oxygen, then also 7gm of calcium oxide will be formed because chemical reactions follows law of constant proportion. As a result, 18gm of oxygen will be left unreacted.
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Question 23 Marks
What is the significance of the symbol of an element? Explain with the help of an example.
Answer
Significance of symbol of element:
  1. It represents name of the element.
  2. It represents one atom of the element.
  3. It represents a definite mass of the element.
  4. It represents one mole of atoms of the element. 
For example: C represents one atom of the element Carbon. It also represents 12gms of Carbon.
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Question 33 Marks
What is the number of zinc atoms in a piece of zinc weighing 10g?
(Atomic mass of Zn = 65u)
Answer
1 mole of Zn = 65gm of zinc = 6.022 × 1023 atoms of zinc
Given mass of zinc = 10gm
No. of moles of zinc $=\frac{10}{65}=0.15\text{ moles of zinc}$
Total no. of atoms in 0.15 moles = 0.15 × 6.022 × 1023 atoms of zn
= 9.264 × 1022 atoms of zn
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Question 43 Marks
The formula of carbonate of a metal M is M2CO3.
  1. What will be the formula of its iodide?
  2. What will be the formula of its nitride?
  3. What will be the formula of its phosphate?
Answer
Let us first work out the valency of the metal M (we know that carbonate ion has a valency of -2):

$\text{Element/ Ion}\ \ \ \ \ \ \ \text{M}\ \ \ \ \ \ \text{CO}^{2-}_3\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \ 2$

It is clear that M has a valency of 1

  1. Formula of iodide of M:

$\text{Element}\ \ \ \ \ \ \ \text{M}\ \ \ \ \ \ \ \text{I}\\ \text{Valency}\ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ 1$

Crossing the valencies, formula of the iodide of M is Ml

  1. Formula of nitride of M:

$\text{Element/ Ion}\ \ \ \ \ \ \ \text{M}\ \ \ \ \ \ \ \ \ \ \ \text{N}^{3-}\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ -3$

Crossing the valencies, formula of the nitride of M is M3N

  1. Formula of phosphate of M:

$\text{Element/ Ion}\ \ \ \ \ \ \ \text{M}\ \ \ \ \ \ \ \ \ \text{PO}^{3-}_4\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \ -3$

Crossing the valencies, formula of the phosphate of M is M3PO4

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Question 53 Marks
Show by means of calculations that 5 moles of CO2 and 5 moles of H2O do not have the same mass. How much is the difference in their masses?
Answer
Molar mass of CO2 = 44g
Molar mass of H2O = 18g
Mass of 5 mole of H2O = 5 × 18g = 90g
Mass of 5 mole of CO2 = 5 × 44g = 220g
So, 5 mole of H2O and 5 mole of CO2 do not have same mass.
And the difference in their masses = 220g - 90g = 130g
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Question 63 Marks
Potassium chlorate decomposes, on heating, to form potassium chloride and oxygen. When 24.5g of potassium chlorate is decomposed completely, then 14.9g of potassium chloride is formed. Calculate the mass of oxygen formed. Which law of chemical combination have you used in solving this problem?
Answer
$\text{Potassium chlorate}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }\text{Potassium chloride}\ +\ \text{Oxygen}\\ \ \ \ \ \ \ \ \ \ \ ^{(24.5\text{gm})}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{(14.9\text{gm})}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{(\text{x gm})}$
Let, x gm of oxygen is formed,
Then, according to law of conservation of mass,
24.5gm = 14.9gm + x gm
So, x = (24.5 - 14.9)gm = 9.6gm.
Thus, 9.6 gm of oxygen is formed in the reaction.
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Question 73 Marks
Magnesium and oxygen combine in the ratio of 3 : 2 by mass to form magnesium oxide. What mass of oxygen gas would be required to react completely with 24h of magnesium?
Answer
$\text{Magnesium}+\text{Oxygen}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ }\text{Magnesium oxide}\\ \ \ \ \ \ \ \ \ 3\text{x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\text{x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{x}$
i.e. three equivalents of Mg reacts with 2 equivalents of O2 to form 1 equivalent of MgO.
When mass of Mg = 3x = 24gm
So, x = 8gm
Then, mass of oxygen required = 2x = 16gm
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Question 83 Marks
If the aluminium salt of an anion X is Al2X3, what is the valency of X? What will be the formula of the magnesium salt of X?
Answer
Valency X in the aluminium salt Al2X3, can be worked out as follows (we know that valency of Al is 3):

$\text{Element}\ \ \ \ \ \ \ \text{Al}\ \ \ \ \ \ \text{X}\\ \text{Valency}\ \ \ \ \ \ \ \ 3\ \ \ \ \ \ \ \ 2$

Thus, it is clear that the valency of X is two.

Formula for the magnesium salt of X can be worked out as follows:

$\text{Element}\ \ \ \ \ \ \ \text{Mg}\ \ \ \ \ \ \text{X}\\ \text{Valency}\ \ \ \ \ \ \ \ 2\ \ \ \ \ \ \ \ \ 2$

Crossing the valencies gives us the formula of the magnesium salt of X i.e., MgX.

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Question 93 Marks
How many moles of oxygen atoms are present in one mole of the following compounds?
  1. Al2O3
  2. CO2
  3. Cl2O7
  4. H2SO4
  5. Al2(SO4)3
Answer
Moles of oxygen atom are:
  1. Al2O3: 3 mole
  2. CO2: 2 mole
  3. Cl2O7:  7 mole
  4. H2SO4: 4 mole
  5. Al2(SO4)3: 12 mole
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Question 103 Marks
Copper sulphate reacts with sodium hydroxide to form a blue precipitate of copper hydroxide and sodium sulphate. In an experiment, 15.95g of copper sulphate reacted with 8.0g of sodium hydroxide to form 9.75g of copper hydroxide and 14.2g of sodium sulphate. Which law of chemical combination is illustrated by this data? Give reason for your choice.
Answer
$\text{CuSO}_4\ +\ 2\text{NaOH}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ }\text{Cu}(\text{OH})_2\ +\ \text{Na}_2\text{SO}_4\\ ^{(15.95\text{gm})} \ \ \ \ \ \ \ \ ^{(8\text{gm})}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{9.75\text{gm}}\ \ \ \ \ \ \ \ \ \ \ \ \ ^{14.2\text{gm}}$
Clearly, in this case,
Total mass of reactants = (15.95gm + 8gm) = 23.95gm
Total mass of products = (9.75gm + 14.2gm) = 23.95gm
Hence, Law of conservation of mass is valid here.
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Question 113 Marks
Calculate the number of moles present in a drop of chloroform (CHCl3) weighing 0.0239g.
(Atomic masses: C = 12u; H = 1u; Cl = 35.5u)
Answer
Given mass of CHCl3 = 0.0239g
Molar mas of CHCl3 = 1 × C + 1 × H + 3 × Cl = 119.5g
No. of moles $=\frac{\text{Given mass}}{\text{Molar mass}}$
No. of moles $=\frac{0.0239}{119.5}=0.0002$
So, no. of molecules present in 0.0239g in chloroform = 0.0002 × 6.022 × 1023
= 12.044 × 1019 molecules
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Question 123 Marks
Calculate the number of molecules in 4g of oxygen.
Answer
32g of oxygen (1 mole of oxygen) has = 6.022 × 1023 molecules of oxygen
So, 4g of oxygen will have $=\frac{6.022\ \times\ 10^{23}\ \times\ 4}{32}$
$=7.528\times10^{22}\text{ molecules of oxygen}$
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Question 133 Marks
Calculate the molecular masses of the following:
  1. Hydrogen, H2
  2. Oxygen, O2
  3. Chlorine, Cl2
  4. Ammonia, NH3
  5. Carbon dioxide, CO2
(Atomic masses: H = 1 u; O = 16 u; Cl = 35.5 u; N = 14 u; C = 12 u)
Answer

  1. Molecular mass of Hydrogen (H2) = 2 × H = 2 × 1u = 2u

  2. Molecular mass of oxygen (O2) = 2 × O = 2 × 16u = 32u

  3. Molecular mass of chlorine (Cl2) = 2 × Cl = 2 × 35.5 = 71u

  4. Molecular mass of Ammonia (NH3) = 1 × N + 3 × H = 14 + 3 = 17u

  5. Molecular mass of carbon dioxide (CO2) = 1 × C + 2 × O = 12 + 32 = 44u

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Question 143 Marks
Calculate the molecular masses of the following compounds:
  1. Methane, CH4
  2. Ethane, C2 H6
  3. Ethene, C2 H4
  4. Ethyne, C2 H2
(Atomic masses: C = 12u; H = 1u)
Answer

  1. Molecular mass of methane (CH4) = 12 + 4 = 16u

  2. Molecular mass of ethane (C2H6) = 2 × 12 + 6 × 1 = 30u

  3. Molecular mass of ethane (C2H4) = 2 × 12 + 4 × 1 = 28u

  4. Molecular mass of ethyne (C2H2) = 2 × 12 + 2 × 1 = 26u

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Question 153 Marks
An element A forms an oxide A2O5.
  1. What is the valency of element A?
  2. What will be the formula of chloride of A?
Answer

  1. Element A forms an oxide A2O5. Crossing the valencies, we can see that the valency of O(oxide) is -2 and that of element A is 5.

  2. Formula of chloride of A:

$\text{Element/ Ion}\ \ \ \ \ \ \ \text{A}\ \ \ \ \ \ \text{Cl}^-\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 5\ \ \ \ -1$

Formula of the chlcride of element A can be worked out by crossing over the valencies. Thus, the formula is ACl5

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