Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6

Q.

25 ml of a solution of Na2CO3 having a specific gravity of 1.25 gram ml required 32.9 ml of a solution of HCl containing 109.5 gram of the acid per litre for complete neutralization. Calculate the volume of 0.84 N H2SO4 that will be completely neutralized by 125 gram of Na2CO3 solution.

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

470 ml

b

400 ml

c

380 ml

d

455 ml

answer is A.

(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

Since Na2CO3  is completely neutralized by HCl 

Also, given molarity of HCl 109.536.5=3

Therefore, milliequivalents of Na2CO3 = milliequivalents of HCl 

N×25 = 32.9×3 N = 3.948

Now, a fresh Na2CO3 solution reacts with H2SO4 

Volume of 125 gm of  Na2CO3 1251.25 = 100 ml

Again we can write,

Milliequivalents of  Na2CO3 = Milliequivalents of H2SO4 

100×3.948 = 0.84×V V =470 ml

Watch 3-min video & get full concept clarity

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon