When 800 mL of 0.5 M nitric acid is heated in a beaker, its volume is reduced to half and 11.5 g of nitric acid is evaporated. The molarity of the remaining nitric acid solution is $x \times 10^{- 2} M$ . (Nearest Integer)
(Molar mass of nitric acid is 63 g $m o l^{- 1}$ )

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

answer is 55.

(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 nitric acid $= 800 \times 10^{- 3} \times 0.5 = 0.4$
Moles of nitric acid evaporated $= \frac{11.5}{63} = 0.182$
Moles left = 0.4 - 0.182 = 0.218
Molarity $= \frac{0.218 \times 10^{3}}{400} = 0.545 M = 54.5 \times 10^{- 2} M$

Watch 3-min video & get full concept clarity

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

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!