Q.

Sulphuric acid reacts with sodium hydroxide as follows H2SO4+2NaOH⟶Na2SO4+2H2OWhen 1 L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained are

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

3.55 g, 0.1M

b

7.10 g,0.025 M

c

5.33 g,0.01M

d

5.33 g,0.25M

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Moles of H2SO4=1L×0.1M=0.1molH2SO41 mol+2 NaOH2 mol→Na2SO41 mol+2H2O  0.1 mol   0.1 mol       0.05mol∴ Molarity mass of Na2SO4=0.05×142=7.1g∴ Molar mass of Na2SO4=142 Molarity = moles of Na2SO4 volume of solution =0.051+1=0.025 mol L−1
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
whats app icon