When 9.3 gm of aniline is reacted with acetic anhydride, mass of acetanilide obtained is m gm, find out value of 2m.

The correct answer is 27.

To find the mass of acetanilide obtained and the value of 2m, we can write down the chemical equation for the acetylation reaction of aniline with acetic anhydride:

$C_6{H}_{5}N{H}_2+(C{H}_{3}CO{)}_{2}O\to C_6{H}_{5}NHCOC{H}_3+C{H}_{3}COOH$

From the balanced chemical equation, 1 mole of aniline reacts with 1 mole of acetic anhydride to yield 1 mole of acetanilide.

Step 1: Calculate the molar masses of the reactants and products:
Molar mass of aniline ($C_6{H}_{7}N$ or $C_6{H}_{5}N{H}_2$):
$(6\times 12)+(7\times 1)+14=72+7+14=93\text{g/mol}$

Molar mass of acetanilide ($C_8{H}_{9}NO$):
$(8\times 12)+(9\times 1)+14+16=96+9+14+16=135\text{g/mol}$

Step 2: Determine the mass of acetanilide obtained (m):
Since 93 g of aniline produces 135 g of acetanilide, we can set up a simple proportion to find the mass of acetanilide ($m$) produced from 9.3 g of aniline:
$m=\frac{135}{93}\times 9.3\text{g}$

$m=13.5\text{g}$

Step 3: Calculate the value of 2m:
Using the value of $m=13.5\text{g}$, we find:
$2m=2\times 13.5=27$