JEE MAIN - Chemistry (2020 - 7th January Evening Slot - No. 4)
The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by :
200 ml of 0.02 N HCl
100 ml of 0.2 N HCl
100 ml of 0.1 HCl
200 ml of 0.4 N HCl
Explanation
NH2CONH2 $$ \to $$ NH3
Using Principle of Atom Conservation
2 $$ \times $$ moles of urea = 1 $$ \times $$ moles of NH3
$$ \Rightarrow $$ 2 $$ \times $$ $${{0.6} \over {60}}$$
$$ \Rightarrow $$ moles of NH3 = 0.02
Also moles of NH3 = moles of HCl, because they react in 1 : 1 ratio.
100 ml of 0.2 N HCl = $${{100 \times 0.2} \over {1000}}$$ = 0.02 mole of HCl
So option (B) is correct.
Using Principle of Atom Conservation
2 $$ \times $$ moles of urea = 1 $$ \times $$ moles of NH3
$$ \Rightarrow $$ 2 $$ \times $$ $${{0.6} \over {60}}$$
$$ \Rightarrow $$ moles of NH3 = 0.02
Also moles of NH3 = moles of HCl, because they react in 1 : 1 ratio.
100 ml of 0.2 N HCl = $${{100 \times 0.2} \over {1000}}$$ = 0.02 mole of HCl
So option (B) is correct.
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