Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The ammonia  released on quantitative reaction of 0.6g, urea <img alt="(NH_{2}CONH_{2})" src="http://entrancecorner.oncodecogs.com/gif.latex?%28N

The ammonia (NH_{3}) released on quantitative reaction of 0.6g, urea (NH_{2}CONH_{2}) with sodium hydroxide (NaOH) can be neutralised by:
Option: 1 200 ml of 0.2 N HCl
Option: 2200 ml of 0.4 N HCl
Option: 3100 ml of 0.1N HCl
Option: 4100 ml of 0.2N HCl
 

Answers (1)

best_answer

2 × mole of Urea = mole of NH_{3} ........(1)
mole of NH_{3} = mole of HCl ........(2)
\therefore mole of HCl = 2 × mole of Urea

mole of HCl =2\times \frac{0.6}{60}=0.02mol ...(i)

[We know , mole = M X V = N X n X V]

 \mathrm{100 \ ml\times 0.2N\times 1=0.02\ mol} ...as (i).

Therefore, Option(4) is correct.

Posted by

Ritika Jonwal

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 registration ends tomorrow; changes introduced in exam this year
anam-khan

BTech Admission 2025 Live: JEE Mains registrations closing tomorrow; VITEEE, MET exam dates out
anam-khan

JEE Main 2025 Live: NTA to close JEE registration on Friday; apply at jeemain.nta.nic.in, correction window

Latest Question