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SECTION 3
Q. No 37 - 40 carry 4 marks each and 2 marks is deducted for every wrong answer
This section contains TWO paragraphs
Based on each paragraph, there will be TWO questions
Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is(are) correct
PARAGRAPH 1

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC  as measured.
(Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1)
Q.
Enthalpy of dissociation (in kJ mol-1) of acetic acid obtained from the Expt. 2 is
  • a)
    1.0
  • b)
    10.0
  • c)
    24.5
  • d)
    51.4
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
SECTION 3Q. No 37 - 40 carry 4 marks each and 2 marks is deducted for ...
HCl + NaOH→NaCl + H2O
n = 100 x1 = 100 m mole = 0.1 mole
Energy evolved due to neutralization of HCl and NaOH = 0.1 x 57 = 5.7 kJ = 5700 Joule
Energy used to increase temperature of solution = 200 x 4.2 x 5.7 = 4788 Joule
Energy used to increase temperature of calorimeter = 5700 – 4788 = 912 Joule
ms.Δt = 912
m.sx5.7 = 912
ms = 160 Joule/oC [Calorimeter constant]
Energy evolved by neutralization of CH3COOH and NaOH
= 200x 4.2x5.6 +160x5.6 = 5600 Joule
So energy used in dissociation of 0.1 mole CH3COOH = 5700 - 5600 = 100 Joule
Enthalpy of dissociation = 1 kJ/mole
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SECTION 3Q. No 37 - 40 carry 4 marks each and 2 marks is deducted for every wrong answerThis section contains TWO paragraphsBased on each paragraph, there will be TWO questionsEach question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these fouroption(s) is(are) correctPARAGRAPH 1When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC as measured. (Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1)Q.Enthalpy of dissociation (in kJ mol-1) of acetic acid obtained from the Expt. 2 isa)1.0b)10.0c)24.5d)51.4Correct answer is option 'A'. Can you explain this answer?
Question Description
SECTION 3Q. No 37 - 40 carry 4 marks each and 2 marks is deducted for every wrong answerThis section contains TWO paragraphsBased on each paragraph, there will be TWO questionsEach question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these fouroption(s) is(are) correctPARAGRAPH 1When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC as measured. (Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1)Q.Enthalpy of dissociation (in kJ mol-1) of acetic acid obtained from the Expt. 2 isa)1.0b)10.0c)24.5d)51.4Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about SECTION 3Q. No 37 - 40 carry 4 marks each and 2 marks is deducted for every wrong answerThis section contains TWO paragraphsBased on each paragraph, there will be TWO questionsEach question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these fouroption(s) is(are) correctPARAGRAPH 1When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC as measured. (Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1)Q.Enthalpy of dissociation (in kJ mol-1) of acetic acid obtained from the Expt. 2 isa)1.0b)10.0c)24.5d)51.4Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for SECTION 3Q. No 37 - 40 carry 4 marks each and 2 marks is deducted for every wrong answerThis section contains TWO paragraphsBased on each paragraph, there will be TWO questionsEach question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these fouroption(s) is(are) correctPARAGRAPH 1When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC as measured. (Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1)Q.Enthalpy of dissociation (in kJ mol-1) of acetic acid obtained from the Expt. 2 isa)1.0b)10.0c)24.5d)51.4Correct answer is option 'A'. Can you explain this answer?.
Solutions for SECTION 3Q. No 37 - 40 carry 4 marks each and 2 marks is deducted for every wrong answerThis section contains TWO paragraphsBased on each paragraph, there will be TWO questionsEach question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these fouroption(s) is(are) correctPARAGRAPH 1When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC as measured. (Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1)Q.Enthalpy of dissociation (in kJ mol-1) of acetic acid obtained from the Expt. 2 isa)1.0b)10.0c)24.5d)51.4Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of SECTION 3Q. No 37 - 40 carry 4 marks each and 2 marks is deducted for every wrong answerThis section contains TWO paragraphsBased on each paragraph, there will be TWO questionsEach question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these fouroption(s) is(are) correctPARAGRAPH 1When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC as measured. (Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1)Q.Enthalpy of dissociation (in kJ mol-1) of acetic acid obtained from the Expt. 2 isa)1.0b)10.0c)24.5d)51.4Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of SECTION 3Q. No 37 - 40 carry 4 marks each and 2 marks is deducted for every wrong answerThis section contains TWO paragraphsBased on each paragraph, there will be TWO questionsEach question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these fouroption(s) is(are) correctPARAGRAPH 1When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC as measured. (Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1)Q.Enthalpy of dissociation (in kJ mol-1) of acetic acid obtained from the Expt. 2 isa)1.0b)10.0c)24.5d)51.4Correct answer is option 'A'. Can you explain this answer?, a detailed solution for SECTION 3Q. No 37 - 40 carry 4 marks each and 2 marks is deducted for every wrong answerThis section contains TWO paragraphsBased on each paragraph, there will be TWO questionsEach question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these fouroption(s) is(are) correctPARAGRAPH 1When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC as measured. (Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1)Q.Enthalpy of dissociation (in kJ mol-1) of acetic acid obtained from the Expt. 2 isa)1.0b)10.0c)24.5d)51.4Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of SECTION 3Q. No 37 - 40 carry 4 marks each and 2 marks is deducted for every wrong answerThis section contains TWO paragraphsBased on each paragraph, there will be TWO questionsEach question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these fouroption(s) is(are) correctPARAGRAPH 1When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC as measured. (Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1)Q.Enthalpy of dissociation (in kJ mol-1) of acetic acid obtained from the Expt. 2 isa)1.0b)10.0c)24.5d)51.4Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice SECTION 3Q. No 37 - 40 carry 4 marks each and 2 marks is deducted for every wrong answerThis section contains TWO paragraphsBased on each paragraph, there will be TWO questionsEach question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these fouroption(s) is(are) correctPARAGRAPH 1When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol-1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC as measured. (Consider heat capacity of all solutions as 4.2 J g-1 K-1 and density of all solutions as 1.0 g mL-1)Q.Enthalpy of dissociation (in kJ mol-1) of acetic acid obtained from the Expt. 2 isa)1.0b)10.0c)24.5d)51.4Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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