Chemistry, asked by raniroy2002, 2 months ago

chemistry MCQ please solve

1. The primary standard solution is-

K2Cr207
KMnO4
HCl
NaOH
2. Phenolphthalein indicator is used for titration of

CH3COOH VS Na2CO3
CH3COOH VS NaOH
HNO3 VS NH4OH
H2SO4 VS Na2CO3
3. Equivalent weight of H2SO4 is

98
49
50
106
4. The pH of (N/10) HCl is

0
1
10
2
5. For the titration of HCl and NH4OH, the suitable indicator is

Methyl Orange
EBT
Phenolphthalein
None of the above
6. Oxidation number of Mn in KMnO4 is

-1
+7
+6
+5
7. The secondary standard solution is

HCl
Na2CO3
K2Cr2O7
All of the above
8. The oxidation number of Cr in K2Cr2O7 is +6

True
False
9. Which one always acts as oxidizing agent?

HNO3
MnO2
H2O2
SO2
10. Which one always acts as reducing agent?

H2O2
H2S
HNO3
SO2​

Answers

Answered by sudipsarkarStreamer
2

Answer:

  • c
  • a
  • d
  • b
  • r
  • a
  • c
  • d
  • b
  • a is your answer

Explanation:

pls mark as braiblist

Answered by Anonymous
2

Answers:

1) The primary standard solution is K2Cr2O7.

=> K2Cr2O7 solution is an oxidizing agent that does not react very easily with organic compounds even in small quantities, it is available in pure state and can be weighed and used easily and is reduced to give green Cr3+ ions this makes it suitable for purposes such as Stoichiometry and analysis. Other examples: H2C2O4, Na2CO3.

2) CH3COOH and NaOH can easily be titrated with phenolphthalein indicator.

=> NaOH is a strong base and CH3COOH is a weak acid, NaOH easily neutralizes CH3COOH and the pH reaches 8 quickly on adding a small quantity of NaOH, phenolphthalein indicator turns pink between 8.2 and 10 pH, thus the end point can easily be indicated, hence phenolphthalein can be used for this titration.

3) 49 is the equivalent weight of H2SO4.

=>Molar mass of H2SO4 = 2 x 1 + 32 + 4 x 16 => 98 grams per mole.

=> Equivalent weight of an acid= molar mass/N-Factor.

=>N-Factor = Basicity of H2SO4 = 2

=> Eq. wt. = 98/2 => 49

4) The pH of N/10 HCl is 1.

=>Normality = 0.1 N

=> Basicity of HCl = 1 (1 H+ ion per molecule of HCl in aqueous state).

=> Normality = Molarity x Basicity of acid

=> 0.1 = Molarity x 1 => Molarity = 0.1 M

=>∵ HCl is monobasic.

=>So, the molarity of H+ for 0.1 M HCl will be 0.1 M.

=> pH = -log₁₀H+ =>pH = -log₁₀0.1 => -1 x -1 => 1

=> Hence, the pH of the solution is 1.

5) Methyl Orange Indicator is best used for titration of HCl and NH4OH.

=> Hydrochloric acid is a strong acid and NH4OH is a weak base, the equivalence point will be below 7 (around 5), as HCl is a very strong acid.

=> Methyl orange turns red at pH 3.1-4.4, thus it can easily indicate when the equivalence point has been reached in a titration involving a strong acid and a weak base.

6) Oxidation number of Mn in KMnO4 is +7.

=> Let the oxidation number be x .

=> +1 + x + 4 x -2 = 0

=> 1 + x -8 = 0

=> x -7 = 0

=> x = +7.

7) The secondary standard solution is HCl.

=>A secondary standard solution is a solution whose concentration is measured not from the number of moles of solute dissolved but from titration with a primary standard solution. EG: KMnO4, NaOH.

8) True, the oxidation number of Cr in K2Cr2O7 is +6.

=> Let oxidation number of Cr in K2Cr2O7 be x.

=> So, 2 * +1 + 2*x + 7*-2 = 0

=> 2 + 2x -14 = 0

=> 2x - 12 = 0

=>2x = 12

=> x = 12/2

=> x = +6

9) HNO3 (Nitric Acid) always acts as an oxidizing agent and never a reducing agent.

=> It gives nascent oxygen on its decomposition along with nitric oxide and water.

=> The nascent oxygen is highly oxidizing in nature and immediately oxidizes both metals and nonmetals readily.

10) H2S (Hydrogen sulphide) always acts as a reducing agent.

=> Hydrogen Sulphide has sulphr in the oxidation state -2 .

=>In this situation, S-2 can only be oxidized by loss of two electons two sulphur .

=> Thus Hydrogen sulphide acts as a reducing agent.

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