The ammonia (.NH3.) released on quantitative reaction of 0.6 g urea (.NH2CONH2.) with sodium hydroxide (NaOH) can be neutralized by

Question

The ammonia (.NH3.) released on quantitative reaction of 0.6 g urea (.NH2CONH2.) with sodium hydroxide (NaOH) can be neutralized by (A) 200 ml of 0.4 N HCl (B) 200 ml of 0.2 N HCl (C) 100 ml of 0.1 N HCl (D) 100 ml of 0.2 N HCl

Tardigrade Admin · Accepted Answer

2× moles of NH3 = moles of urea moles of NH3= moles of HCl (100 × 0.2/1000)= moles of HCl =0.02 mole of HCl

Q. The ammonia $(N H_{3})$ released on quantitative reaction of 0.6 g urea $(N H_{2} CON H_{2})$ with sodium hydroxide (NaOH) can be neutralized by

200 ml of 0.4 N HCl

200 ml of 0.2 N HCl

100 ml of 0.1 N HCl

100 ml of 0.2 N HCl

$2 \times m o l es o f N H_{3}$ = moles of urea
moles of $N H_{3} =$ moles of HCl
$\frac{100 \times 0.2}{1000} =$ moles of HCl $= 0.02$ mole of HCl

Q. The ammonia (NH3​) released on quantitative reaction of 0.6 g urea (NH2​CONH2​) with sodium hydroxide (NaOH) can be neutralized by