Purification and Characteristics of Organic Compounds MCQs for NEET — Chemistry Questions with Answers

(d) It is the process to prepare soda extract.

(d) Steam distillation is used to purify the substances which

(i) are volatile in steam but are immiscible with water.

(ii) possess sufficiently high vapour pressure at the boiling point of water.

(iii) contain non- volatile impurities.

The process of steam distillation can also be used to separate a mixture of two organic compounds one of which is steam volatile while the other is not. In ortho and para-nitrophenols, the latter is non-volatile, hence they are separated by steam distillation.

$$\begin{gathered} \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2} \left(Here, P_1=715-15=700 mm\right) \\ \frac{700\times 55}{300}=\frac{760\times V_2}{273} \\ {V}_2=46 ml at NTP \\ \% of N=\frac{28}{22400}\times \frac{volume of{N}_2 at NTP\left(in ml\right)}{weight of organic compound \left(in g\right)}\times 100 \\ =\frac{28}{22400}\times \frac{46}{0.35}\times 100=16.45 \end{gathered}$$

Actually, ${\mathrm{Na}}_2\mathrm{S}$ & NaCN are removed as ${\mathrm{H}}_2\mathrm{S}$ & HCN by conc. ${\mathrm{HNO}}_3$

$$\begin{gathered} {\mathrm{Na}}_2\mathrm{S}+2{\mathrm{HNO}}_3\to 2{\mathrm{NaNO}}_3+{\mathrm{H}}_2\mathrm{S} \\ \mathrm{NaCN}+{\mathrm{HNO}}_3\to {\mathrm{NaNO}}_3+\mathrm{HCN} \end{gathered}$$

When the Lassaigne's extract is boiled with concentrated HNO3, it decomposes Na2S and NaCN (if formed during the test) to prevent their interference in the detection of halogens. This step ensures that the precipitation of AgCl is due to the presence of halogens in the given organic compound.

Ether Boiling point = 34.5° C

Benzene boiling point = 80°C 

Both can be separated by Distillation.

10 ml of $1M\; H2SO4$ corresponds to 10 mmol.

(C) NH₂-NH₂ (inorganic compound give -ve Lassaigne's test)

Mass of the compound = 0.2

Mass of CO₂ formed = 0.44

Percentage formed = 0.44

Percentage of

$$\begin{gathered} \mathrm{C}=\frac{12}{44}\times \frac{\mathrm{Mass} \mathrm{of} {\mathrm{CO}}_2 \mathrm{formed}}{\mathrm{Mass} \mathrm{of} \mathrm{QC}} \times 100 \\ = \frac{12}{44}\times \frac{0.44}{0.2}\times 100 \\ = 60\% \end{gathered}$$

When N + S coexist, NaSCN forms and reacts with Fe³⁺ → blood-red $[\text{Fe(SCN)}]^{2+}$.