Chemistry Test - 23

Factors affecting electrolytic conductance are:

Concentration of ions: The sole reason for the conductivity of electrolytes is the ions present in them. The conductivity of electrolytes increases with an increase in the concentration of ions as there will be more charge carriers if the concentration of ions is more and hence the conductivity of electrolytes will be high.

Nature of electrolyte: Electrolytic conduction is significantly affected by the nature of electrolytes. The degree of dissociation of electrolytes determines the concentration of ions in the solution and hence the conductivity of electrolytes.

Substances such as \(\mathrm{CH}_{3} \mathrm{COOH}\), with a small degree of separation, will have less number of ions in the solution and hence their conductivity will also below, and these are called weak electrolytes. Strong electrolytes such as

\(\mathrm{KNO}_{3}\) have a high degree of dissociation and hence their solutions have a high concentration of ions and so they are good electrolytic conductance.

Temperature: Temperature affects the degree to which an electrolyte gets dissolved in solution. It has been seen that higher temperature enhances the solubility of electrolytes and hence the concentration of ions which results in an increased electrolytic conduction.

When water is added to an aqueous solution the number of ions per unit volume decreases i.e., the concentration of ions decreases and hence thereby conductivity gets decreased.

In the reaction of potassium dichromate with hydrogen sulphide in presence of an acidic medium of sulphuric acid, a redox reaction takes place as follows:

During the reaction, the potassium chromate acts as the oxidizing agent, due to the presence of the hexavalent chromium in (+6) oxidation state. It is thus reduced to (+3) oxidation state.

Then, the reduction half-cell reaction will be:

\(\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}+14 \mathrm{H}^{+}+6 e^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\)

Also, the sulphide \(S^{2-}\) acting as the reducing agent, is oxidised to sulphur (S) in the reaction. Then, the oxidation half-cell reaction will be:

\(\mathrm{S^{2-} \longrightarrow S+2 e^{-}}\)

Thus, combining the oxidation and reduction half-cell reaction, the balanced equation obtained is:

\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+3 \mathrm{H}_{2} \mathrm{~S}+4 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 3 \mathrm{S}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}+7 \mathrm{H}_{2} \mathrm{O} \)

The products formed are sulphur, potassium sulphate, chromium sulphate and water.

Glycerol is Trihydric alcohol.

Compounds having one −OH group is called monohydric alcohol.

Compounds having two −OH group is called dihydric alcohol.

Compounds having three −OH group is called trihydric alcohol.

Image of glycerol is given below, it has three −OH group

We know that:

Oxidation number of potassium, \((K)=1\)

And, Oxidation number of oxygen, \((O)=-2\)

For \(K I O_{4}\):

Let lodine has \(x\) oxidation number.

\(1+x+4(-2)=0\)

\(x=8-1\)

\(x=7\)

For \(K M n O_{4}\):

Let \(M n\) has\(a\)oxidation number.

\(1+a+4(-2)=0\)

\(a=7\)

The given reaction is \(a 1,4\) addition reaction.

In the first step, a molecule of \(H B r\) is added to a conjugated diene. Both \(H\) atom and \(B r\) atom are added to terminal \(C\) atoms. This is \(1,4\) addition or conjugate addition. In the next step, the reaction with \(M g\) in presence of ether gives a Grignard reagent \((R-M g X)\) which then attacks a molecule of carbon dioxide. Acid hydrolysis gives carboxylic acid.

Mixture of \(\mathrm{\left[ Conc~ HNO _3\right.}\) conc \(\mathrm{\left.H _2 SO _4\right]}\) gives \(\mathrm{{ }^{+} NO _2}\) which acts as electrophile and in nitrobenzene \(\mathrm{- NO_2}\) group is a meta directing group so nitronium ion attacks at meta position.

Preparation of Alkenes from Alkyl Halides:

Alkyl halides, upon heating with an alcoholic solution of potassium hydroxide, forms alkenes and halogen acids as by-products. This reaction is known as dehydrohalogenation.

Since hydrogen is removed from \(\beta\) carbon, this reaction is also termed as \(\beta\) elimination.

Saytzeff Rule

In dehydrohalogenation reactions, the major product is alkene that has a maximum number of alkyl groups attached to doubly bonded carbon and other products are considered to be minor.

Hence, the total number of alkenes formed is 5.

In the manufacturing of sulphuric acid, a step involves the reaction between SO₂ and O₂ in the presence of platinized asbestos to give SO₃.

SO₂ + O₂ → SO₃

​In the above reaction if the traces of arsenious oxide (As₂O₃) is used then it acts as a catalytic poison by decreasing the activity of the reaction.