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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)


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} (i).

Therefore, Option(4) is correct.

Posted by

Ritika Jonwal

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