When 400 mL of 0.2 M H₂SO₄ solution is mixed with 600 mL of of 0.1 M NaOH solution the increase in temperature of the final solution is _____ *10⁻² K. (Round off to the Nearest Integer).
[Use : H⁺(aq) + OH⁻(aq) → H₂O : ΔₙH = –57.1kJmol⁻¹
Specific heat of H₂O = 4.18JK⁻¹g⁻¹
Density of H₂O = 1.0gcm⁻³
Assume no change in volume of solution on mixing]

33 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
A
5 months ago

H? SO? + 2NaOH → Na? SO? + 2H? O
Milli mole: 80 60 (limiting reagent)
H? + OH? → H? O
Milli mole: 160 60
Milli mole: 160-60 0 60
Total heat produced = (60/1000) x 57.1 x 1000 = 3426 J
Total heat absorbed = msΔT
ΔT = Totalheat / (m x s)
= 3426J / (1000 x 4.18) = 0.819K
= 81.9 x 10? ² K
Ans. = 82 (the nearest)

Thumbs Up IconUpvote Thumbs Down Icon

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers

Learn more about...

Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight 2025

Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight 2025

View Exam Details

Most viewed information

Summary

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Have a question related to your career & education?

or

See what others like you are asking & answering