Question

$1.86\mathrm{g}$ of aniline completely reacts to form acetanilide. $10\%$ of the product is lost during purification. The amount of acetanilide obtained after purification $\left(\mathrm{in}\mathrm{g}\right)$ is $\mathrm{\_\_\_\_}\times {10}^{–2}$.

Answer 1

Step 1: Given data:

The weight of Aniline $\left({\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\right)$ is ${\mathrm{w}}_{\left({\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\right)}=1.86\mathrm{g}$

Step 2: Find the Molecular weights of the reactant and product:

$\underset{Reac\tan t}{{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2}\to \underset{\mathrm{Product}}{{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NHCOCH}}_3}$

Molecular weight of Aniline is $\mathrm{GMM}=93\mathrm{g}$

Molecular weight of Acetanilide is $\mathrm{GMM}=135\mathrm{g}$

Step 3: Find the remaining weight of Acetanilide:

$93\mathrm{g}$ of Aniline produces $135\mathrm{g}$ of Acetanilide

$1.86\mathrm{g}$ of Aniline produce $?$

Weight of Acetanilide is ${\mathrm{w}}_{\left({\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NHCOCH}}_3\right)}=\frac{1.86\mathrm{g}\times 135\mathrm{g}}{93\mathrm{g}}$

$\Rightarrow {\mathrm{w}}_{\left({\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NHCOCH}}_3\right)}=2.7\mathrm{g}$

At $10\%$ loss, $90\%$ of Acetanilide has remained after purification.

$$\begin{gathered} \therefore \mathrm{Amount}\mathrm{of}\mathrm{product}\mathrm{obtained}=\frac{2.7\times 90}{100} \\ =2.43\mathrm{g} \end{gathered}$$

Hence, the amount of Acetanilide obtained after the purification is $\mathbf{2}\mathbf{.}\mathbf{43}\mathbf{}\mathbf{g}$.