X g of ethanamine was subjected to reaction with NaNO2/HCl followed by hydrolysis to liberate N2 and HCl. The HCl generated was completely neutralised by 0.2 moles of NaOH. X is g.
Answer
9
Solution
C2H5NH2 NaNO2/HCl C2H5N2+Cl-
C2H5N2+Cl- + H2O C2H5OH + N2 + HCl
∵ 1 mol amine gives 1 mol HCl, and HCl + NaOH NaCl + H2O (1 : 1)
⇒ HCl = 0.2 mol ⇒ ethanamine = 0.2 mol
∴ X = 0.2 × 45 = 9 g (M of C2H5NH2 = 45)
Answer: 9