How are ethylamine and ethyl methyl amine distinguished by using nitrous acid?

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Solution

Ethylamine (a primary amine) reacts with nitrous acid in cold condition to give ethyl alcohol with liberation of nitrogen gas.
$C H_{3} - C H_{2} - N H_{2} + H N O_{2} N a N O_{2} + H C l - ------- \to 273 - 278 K C H_{3} - C H_{2} - O H + H_{2} O + N_{2} ↑$
Ethyl methyl amine (a secondary amine) reacts with nitrous acid in cold condition to give yellow oily N-nitrosoethyl methyl amine.
$C H_{3} - C H_{2} - N H - C H_{3} + H N O_{2} N a N O_{2} + H C l - ------- \to 273 - 278 K C H_{3} - C H_{2} - N (C H_{3}) - N = O + H_{2} O$

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