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Science Question Answer (3 Marks)

Atoms and Molecules · Maharashtra Board STD 9 (MSBSHSE - Semi English Medium). 15 questions.

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Question Answer (3 Marks)

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

Question 13 Marks
Show by means of calculations that $5$ moles of $CO_2​​​​​​​$​​​​​​​ and 5 moles of $H_2O$ do not have the same mass. How much is the difference in their masses?
Answer
Molar mass of $CO _2=44 g$
Molar mass of $H _2 O =18 g$
Mass of 5 mole of $H _2 O =5 \times 18 g=90 g$
Mass of 5 mole of $CO _2=5 \times 44 g=220 g$
So, $5$ mole of $H _2 O$ and $5$ mole of $CO _2$ do not have same mass.
And the difference in their masses $=220 g-90 g=130 g$
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Question 23 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 $24\ h$ 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 $O _2$ to form $1$ equivalent of $MgO$ .When mass of $M g=3 x =24\ gm$
So, $x=8\ gm$
Then, mass of oxygen required $=2 x =16\ gm$
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Question 33 Marks
What is the number of zinc atoms in a piece of zinc weighing $10\ g$?
(Atomic mass of $Zn = 65\ u$)
Answer
$1$ mole of $Zn =65 gm$ of zinc $=6.022 \times 10^{23}$ atoms of zinc
Given mass of zinc $=10 gm$
No. of moles of zinc $=\frac{10}{65}=0.15$ moles of zinc
Total no. of atoms in $0.15$ moles $=0.15 \times 6.022 \times 10^{23}$
atoms of zn $=9.264 \times 10^{22}$ atoms of zn
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Question 43 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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MCQ 53 Marks
Calculate the molecular masses of the following compounds:
(Atomic masses: $C = 12\ u; H = 1\ u$)
  • A
    Methane, $CH_4$
     
  • B
    Ethane, $C_2 H_6$
     
  • C
    Ethene, $C_2 H_4$
     
  • D
    Ethyne, $C_2 H_2$
Answer
  1. Molecular mass of methane $(CH_4) = 12 + 4 = 16u$
  2. Molecular mass of ethane $(C_2H_6) = 2 \times 12 + 6 \times 1 = 30u$
  3. Molecular mass of ethane $(C_2H_4) = 2 \times 12 + 4 \times 1 = 28u$
  4. Molecular mass of ethyne $(C_2H_2) = 2 \times 12 + 2 \times 1 = 26u$
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Question 63 Marks
Calculate the molecular masses of the following:
  1. Hydrogen, $H_2$
  2. Oxygen, $O_2$
  3. Chlorine, $Cl_2$
  4. Ammonia, $NH_3$
  5. Carbon dioxide, $CO_2$
(Atomic masses: $H = 1\ u; O = 16\ u; Cl = 35.5\ u; N = 14\ u; C = 12\ u$)
Answer
Molecular mass of Hydrogen $(H_2) = 2 \times H = 2 \times 1u = 2\ u$
Molecular mass of oxygen $(O_2) = 2 \times O = 2 \times 16u = 32\ u$
Molecular mass of chlorine $(Cl_2) = 2 \times Cl = 2 \times 35.5 = 71\ u$
Molecular mass of Ammonia $(NH_3) = 1 \times N + 3 \times H = 14 + 3 = 17\ u$
Molecular mass of carbon dioxide $(CO_2) = 1 \times C + 2 \times O = 12 + 32 = 44\ u$
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Question 73 Marks
Calculate the number of moles present in a drop of chloroform $\left( CHCl _3\right)$ weighing $0.0239\ g .$
(Atomic masses: $C =12 u ; H =1 u ; Cl =35.5 u$ )
Answer
Given mass of $CHCl _3=0.0239 g$
Molar mas of $CHCl _3=1 \times C +1 \times H +3 \times Cl =119.5 g$
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 0.0239 g in chloroform $=0.0002 \times 6.022 \times 10^{23}$
$=12.044 \times 10^{19}$ molecules
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Question 83 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 93 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 103 Marks
An element A forms an oxide $A_2O_5.$
  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 $A_2O_5$​​​​​​​. 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 $ACl_5$
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Question 113 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 123 Marks
The formula of carbonate of a metal M is $M_2CO_3$.
  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 $M_3N$​​​​​​​
  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 $M_3PO_4​​​​​​​$​​​​​​​
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Question 133 Marks
If the aluminium salt of an anion $X$ is $Al_2X_3$, what is the valency of $X$? What will be the formula of the magnesium salt of $X$?
Answer
Valency X in the aluminium salt $Al_2X_3$, 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 143 Marks
How many moles of oxygen atoms are present in one mole of the following compounds?
  1. $Al_2O_3$
  2. $CO_2$
  3. $Cl_2O_7$
  4. $H_2SO_4$
  5. $Al_2(SO_4)_3$
Answer
Moles of oxygen atom are:
  1. $Al_2O_3$: 3 mole
  2. $CO_2$: 2 mole
  3. $Cl_2O_7$: 7 mole
  4. $H_2SO_4$: 4 mole
  5. $Al_2(SO_4)_3$: 12 mole
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Question 153 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 Answer (3 Marks) - Science STD 9 Questions - Vidyadip