Important Question From NCERT Exemplar Class 10 Science Chapter 1
Some NCERT Exemplar Class 10 Science Chapter 1 Chemical Reactions and Equation questions and answers are given below that will help you tackle complex problems. The questions below will help you evaluate your understanding of the concepts.
Question 1: A dilute ferrous sulphate solution was gradually added to the beaker containing acidified permanganate solution. The light purple colour of the solution fades and finally disappears. Which of the following is the correct explanation for the observation?
(a) $\mathrm{KMnO}_4$ is an oxidising agent, it oxidises $\mathrm{FeSO}_4$
(b) $\mathrm{FeSO}_4$ acts as an oxidising agent and oxidises $\mathrm{KMnO}_4$
(c) The colour disappears due to dilution; no reaction is involved
(d) $\mathrm{KMnO}_4$ is an unstable compound and decomposes in the presence of $\mathrm{FeSO}_4$ to a colorless compound.
Answer:
The purple color of the solution was due to potassium permanganate, which acts as an oxidizing agent. When it oxidises $\mathrm{FeSO}_4$ the purple colour fades away
Hence, the correct option is option (a).
Question 2: A sample of a chromium-containing alloy weighing 3.45 g was dissolved in acid, and all the chromium in the sample was oxidized to $\mathrm{CrO}_4^{2-}$. It was then found that 3.0 g of $\mathrm{Na}_2 \mathrm{SO}_3$ was required to reduce the $\mathrm{CrO}_4^{2-}$ to $\mathrm{CrO}_2^{-}$in a basic solution, with $\mathrm{SO}_3^{2-}$ getting oxidized to $\mathrm{SO}_4^{2-}$. Then the \% of chromium in the alloy is $\qquad$ .
(Round off your answer to the nearest integer)
Answer:
The reaction involved is
$$
2 \mathrm{CrO}_4^{2-4}+3 \mathrm{SO}_3^{2-}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{CrO}_2^{-}+3 \mathrm{SO}_4^{2-}+2 \mathrm{OH}^{-}
$$
Also,
$$
\begin{aligned}
\mathrm{molCrO}_4^{2-} & =\left(3.0 \mathrm{gNa}_2 \mathrm{SO}_3\right) \times\left(\frac{1 \mathrm{molNa}_2 \mathrm{SO}_3}{126 \mathrm{gNa}_2 \mathrm{SO}_3}\right) \times\left(\frac{1 \mathrm{molSO}_3^{2-}}{1 \mathrm{molNa}_2 \mathrm{SO}_3}\right) \times\left(\frac{2 \mathrm{molCrO}_4^{2-}}{3 \mathrm{molSO}_3^{2-}}\right) \\
& =1.58 \times 10^{-2} \mathrm{molCrO}_4^{2-}
\end{aligned}
$$
Since there is one mole of Cr in each mole of $\mathrm{CrO}_4^{2-}$, then the above number of moles $\mathrm{CrO}_4^{2-}$ is also equal to the number of moles of Cr that were present:
$0.0158 \mathrm{molCr} \times 52.00 \mathrm{~g} / \mathrm{mol}=0.825 \mathrm{gCr}$ in the original alloy.
$\%$ of $\mathrm{Cr}=(0.825 \mathrm{gCr} / 3.450 \mathrm{~g}$ sample $) \times 100 \%=23.9 \% \mathrm{Cr}$
Alternately,
gram equiv of $\mathrm{CrO}_4^{2-} \equiv$ gram equiv of $\mathrm{SO}_3^{2-}$
$\Rightarrow\left(\right.$ Moles of Cr or $\left.\mathrm{CrO}_4^{2-}\right) \times 3=\frac{3}{126 / 2}$
$\Rightarrow$ Moles of $\mathrm{Cr}=\frac{1}{63}$
Therefore, $\%$ of Cr in alloy $=\frac{52 / 63}{3.45} \times 100=23.92 \%$
Hence, the answer is (24).
Question 3: Which of the following methods can be used to prevent the food from getting rancid:
i. Storing the food in the air-tight containers
ii. Storing the food in refrigerator
iii. Keeping food in open
iv. Keeping food in moist atmosphere
(1) (i) and (ii)
(2) (i) and (iii)
(3) (iii) and (iv)
(4) All of the above
Answer:
Rancidity can be prevented by storing food in airtight containers because in airtight containers there is little exposure to oxygen then oxidation of fats and oils present in food is slowed down and thus rancidity is retarded. Rancidity can be prevented by keeping food in a refrigerator because in a refrigerator when the food is kept the oxidation of fats and oils in it is slowed down due to low temperature and thus rancidity is retarded.
Hence, the answer is the option (1).
Question 4: The white precipitates formed In a chemical reaction between sulphuric acid and barium chloride solution is of:
(1) Sulphur
(2) Barium sulphate
(3) Chlorine
(4) Silver chloride
Answer:
White precipitates formed In a chemical reaction between sulphuric acid and barium chloride solution is of Barium sulphate.
Hence, the answer is the option (2).
Question 5: Respiration is known as the process of combustion of glucose undergoes with oxygen in the cells of our body to produce energy, is a kind of:
(1) Exothermic process
(2) Reversible process
(3) Endothermic process
(4) Physical process
Answer:
Respiration, which is the process of combustion of glucose is an exothermic reaction.
Hence, the answer is the option (1).
Approach To Solve Class 10 Science Chapter 1 Questions
To solve Class 10 Science NCERT Exemplar Solutions Chapter 1 Chemical Reactions and Equations questions, it is important to follow a systematic approach. It is recommended to strategies your study plan to solve the questions of this chapter. The following are the points that will help you build a good approach.
1. Recall the fundamental concepts like Chemical reaction, Balance the equation, types of chemical reaction, etc. Students can also follow Class 10 Science Chapter 1 Chemical Reactions and Equation notes to understand all these concepts better.
2. Learn about different kinds of reaction
- Combination: $A+B \rightarrow A B$
- Decomposition: $A B \rightarrow A+B$
- Displacement: $A+B C \rightarrow A C+B$
- Double Displacement: $A B+C D \rightarrow A D+C B$
- Redox (Oxidation-Reduction): Identify the species that is oxidized/reduced.
3. Questions from balancing of chemical equations are asked frequently in exams to solve them effectively first convert the word equation into a chemical equation and balance it by ensuring equal atoms of each element on both sides.
4. To identify whether the change is Physical or Chemical, note down the indicators such as change in colour, evolution of gas, precipitate formation, or temperature change to identify reactions practically.
5. Practice again and again, as it will help in mastering NCERT Exemplar Class 10 Science Chapter 1 Chemical Reactions and Equation. Start with simpler problem,s then gradually move to difficult ones. Questions from this chapterare asked directly in the CBSE and State board exams.