Home Maths JEE - Advanced Previous Year Paper JEE-Advanced-2015-Paper-2 SECTION 3 (Maximum Marks: 16) This section c…
Maths JEE - Advanced Previous Year Paper JEE-Advanced-2015-Paper-2 Numeric Response
Published on: August 11, 2026

SECTION 3 (Maximum Marks: 16)

This section contains TWO paragraphs

Based on each paragraph, there will be TWO questions

Each question has FOUR options

A
, 1.0 2.8
B
, 10.0 4.7
C
and 24.5 5.0
D
. ONE OR MORE THAN ONE of these four option(s) is(are) correct Marking scheme: +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened 0 In none of the bubbles is darkened – 2 In all other cases 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 5.7 o C 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 (K a = 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 5.6 o C was 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 ) (i) Enthalpy of dissociation (in kJ mol -1 ) of acetic acid obtained from the Expt. 2 is 51.4 (ii) The pH of the solution after Expt. 2 is 7.0

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Text Solution

Verified by Experts
The correct answer is:
200

(i) 2 HCl + NaOH NaCl + H 2 O

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. t = 912

m.s x 5.7 = 912

ms = 160 Joule/ 0 C [Calorimeter constant]

Energy evolved by neutralization of CH 3 COOH and NaOH

= 200x4.2x5.6x160x5.6=5600 Joule

So energy used in dissociation of 0.1 mole CH3COOH = 5700-5600 = 100 Joule

Enthalpy of dissociation = 1 kJ/mole

(ii)

pH=4.7

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