A) 200 ml of 0.4 N HCl
B) 100 ml of 0.1 N HCl
C) 200 ml of 0.2 N HCl
D) 100 ml of 0.2 N HCl
Correct Answer: D
Solution :
[d] 2 moles of \[N{{H}_{3}}\]will react with 2 mole of HCl. 0.6 g of urea give \[=\frac{0.6}{60}\times 2\,=0.02\]mol of \[N{{H}_{3}}\] \[100\times 0.2\text{ }N\text{ }HCl=0.02\text{ }mol\] of HClYou need to login to perform this action.
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