59 g of an amide obtained from a carboxylic acid, RCOOH, upon heating with alkali liberated 17 g of ammonia. The acid is

Question

59 g of an amide obtained from a carboxylic acid, RCOOH, upon heating with alkali liberated 17 g of ammonia. The acid is (A) Formic acid (B) Acetic acid (C) Propionic acid (D) Benzoic acid.

Tardigrade Admin · Accepted Answer

RCONH2 + NaOH arrow RCOONa + NH3 As 1 mole of NH3 is given by one mole of amide. Thus mol. mass of RCONH2 is A + 12 + 16 + 14 + 2 = 59 or A + 44 = 59 A = 59 - 44 = 15 Hence alkyl group with mol. mass A = 15 is CH3 Thus, acid is CH3COOH .

Q. 59 g of an amide obtained from a carboxylic acid, RCOOH, upon heating with alkali liberated 17 g of ammonia. The acid is

Formic acid

Acetic acid

Propionic acid

Benzoic acid.

$RCON H_{2} + N a O H \to RCOON a + N H_{3}$
As 1 mole of NH3 is given by one mole of amide. Thus mol. mass of $RCON H_{2}$ is
A + 12 + 16 + 14 + 2 = 59 or A + 44 = 59
A = 59 - 44 = 15
Hence alkyl group with mol. mass A = 15 is $C H_{3}$
Thus, acid is $C H_{3} COO H$ .

Q. 59 g of an amide obtained from a carboxylic acid, RCOOH, upon heating with alkali liberated 17 g of ammonia. The acid is

Formic acid

Acetic acid

Propionic acid

Benzoic acid.

RCONH2​+NaOH→RCOONa+NH3​ As 1 mole of NH3 is given by one mole of amide. Thus mol. mass of RCONH2​ is A + 12 + 16 + 14 + 2 = 59 or A + 44 = 59 A = 59 - 44 = 15 Hence alkyl group with mol. mass A = 15 is CH3​ Thus, acid is CH3​COOH .

$RCON H_{2} + N a O H \to RCOON a + N H_{3}$
As 1 mole of NH3 is given by one mole of amide. Thus mol. mass of $RCON H_{2}$ is
A + 12 + 16 + 14 + 2 = 59 or A + 44 = 59
A = 59 - 44 = 15
Hence alkyl group with mol. mass A = 15 is $C H_{3}$
Thus, acid is $C H_{3} COO H$ .