Preparation and Properties of Compounds MCQ Questions & Answers in Inorganic Chemistry | Chemistry

$$pH$$  range for methyl orange is
\[\xleftarrow[\text{Pinkish red}]{}3.9-4.5\xrightarrow[\text{Yellow }]{}\]
Generally, weak bases have $$pH$$ greater than 7. When methyl orange is added to a weak base solution, solution becomes yellow. This solution is then titrated by a strong acid and at the end point pH will be less than 3.1.
∴ Solution becomes pinkish red.

To convert covalent compounds into ionic compounds such as $$NaCN,N{a_2}S,NaX,$$     etc.

The detection of chlorine, sulphur and nitrogen in organic compounds is done by Lassaigne’s test.

$$\% \,\,{\text{ratio of}}\,C:H\,\,{\text{is}}\,\,6:1$$     $${\text{and}}\,\,C:O\,\,{\text{is}}\,\,3:4\,\,{\text{or}}\,\,6:8$$
$$\eqalign{ & \therefore \,\,\% \,\,{\text{ratio of}}\,\,C:H:O\,\,{\text{is}}\,\,6:1:0 \cr & {\text{( Total = 15)}} \cr & \% \,{\text{of}}\,\,C = \frac{6}{{15}} \times 100 = 40 \cr & \frac{{40}}{{12}} = 3.33 = 1 \cr & \% \,\,{\text{of}}\,\,H = \frac{1}{{15}} \times 100 = 6.6 \cr & \frac{{6.6}}{1} = 6.6 = 2 \cr & \% \,\,{\text{of}}\,\,O = \frac{8}{{15}} \times 100 = 53.3 \cr & \frac{{53.3}}{{16}} = 3.3 = 1 \cr & {\text{simple ratio}}:C{H_2}O \cr & \therefore \,\,{\text{The compound is}}\,\,HCHO \cr} $$

Lassaigne’s test is given by only those compounds which contain both carbon and nitrogen. When compounds containing $$C$$  and $$N$$  heated with sodium, then it form $$NaCN$$   which is easily detected by $$FeC{l_3}.$$
Or
Some compounds live hydrazine $$\left( {N{H_2} \cdot N{H_2}} \right)$$    although contain nitrogen but they do not respond Lassaigne’s test because they do not have any carbon and hence, $$NaCN$$   is not formed.

Carboxylic acids are soluble in sodium bicarbonate but phenol are not dissolve in it, so they are separated because carboxylic acid react with $$NaHC{O_3}$$   and form sodium carboxylate.
$$R - COOH + NaHC{O_3} \to $$      $$R - CO{O^ - }N{a^ + } + {H_2}C{O_3}$$

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 $$para$$  - nitrophenol, $$ortho$$  - nitrophenol has intramolecular $$H$$ - bonding. So, it has lower boiling point. Intermolecular $$H$$ - Bonding more strong then intramolecular $$H$$ - bonding. Whereas $$para$$  - nitrophenol has intermolecular $$H$$ - bonding. So, it has higher boiling point. Due to difference in boiling points $$ortho$$  and $$para$$  - nitrophenol can be separated from each other by distillation.

Carboxylic acids dissolve in sodium bicarbonate, while phenol does not.

$$N{a_2}S$$  and $$NaCN,$$  if present in the extract, will be decomposed to $${H_2}S$$  and $$HCN$$  by $$HN{O_3}.$$
$$\eqalign{ & NaCN + HN{O_3} \to NaN{O_3} + HCN \cr & N{a_2}S + 2HN{O_3} \to 2NaN{O_3} + {H_2}S \cr} $$
These will escape from the solution and will not interfere with the test for halogens.

Let the amount of $$N{a_2}C{O_3}$$   present in the mixture be $$x\,g.$$  $$N{a_2}S{O_4}$$  will not react with $${H_2}S{O_4}.$$  Then
$$\eqalign{ & \frac{x}{{53}} = \frac{{20 \times 0.1 \times 10}}{{1000}} \cr & \therefore \,x = 1.06g \cr & \therefore \,{\text{Percentage of}}\,N{a_2}C{O_3} \cr & = \frac{{1.06 \times 100}}{{1.25}} \cr & = 84.8\% \cr} $$