Organic Chemistry Some Basic Principles and Techniques Test - 11

Mass of silver salt taken = 0.228 gm Mass of silver left = 0.162 gm Basicity of acid = 2

Step 1- To calculate the equivalent mass of the silver salt

\({\frac{Eq. mass \,of \,silver \,salt}{Eq. mass \,of\,silver}}\)  = \({\frac{Mass\,of\,Acid\,taken}{Mass\,of\,silver\,left}}\)

= \({\frac{E}{108}}\) = \({\frac{0.228}{0.162}}\times\,108\)

=   152 (Eq.mass of silver salt)

Step 2 - To calculate the eq. mass of acid. Eq. mass of acid = Eq. mass of silver salt - Eq. mass of Ag + Basicity = 152 - 108 + 1 = 152 - 109 = 43 (Eq. mass of acid).

Step 3- To determine the molecular mass of acid. Mol. mass of the acid = Eq. mass of acid ´ basicity = \(45\times2\) = 90.

Solution contain He + \(CH_4\) their mol. wt = 4 +16 =20 % wt of

\(CH_4\) = \({\frac{wt\,of\,CH_4}{Total\, wt}}\times 100\)

= \({\frac{16}{20}}\times100\)

= 80.0

Due to its volatile nature camphor is often used in molecular mass determination.

In Kjeldahl’s method, the nitrogen is estimated in the form of ammonia, which is obtained by heating compounds with

\(NaOH.CH_3CONH_2 + NaOH \overset{\Delta}{\longrightarrow}\) \(CH_3COONa + H_2O + NH_3\)

IUPAC name of acetyl salicylic acid is 2-acetoxy benzoic acid

IUPAC name of \(CH_3-O-C_2H_5\) is Methoxyethane

The IUPAC name of \(CH_3C\equiv N\) is Ethanenitrile

IUPAC name of given compound is 1-pentanal.

\(Cl_2-C=C-Cl_2\) (Hybridization of \(Cl-C-Cl\) is \(sp^2\) and bond angle is 120° and 109.5°)

Minimum mass of sulphur = wt. of its one atom = 32

3.4 gms of sulphur present in 100 gms.

 ∴ 32 gms of sulphur present in  

\({\frac{100\times32}{3.4}}\) = 940

Halogen is estimated by carius method.

Molecular mass = \(2\times V.D\)

= \(2\times 37\)

= 74

Kjeldahl’s method depends upon the fact that most of the organic compounds containing nitrogen are quantitatively decomposed to give \((NH_4)_2SO_4\) when heated strongly with conc. \(H_2SO_4\). In this method \(CuSO_4\) acts as catalytic agent.

Urea \((NH_2CONH_2)\) has molecular wt. 60 and wt. of Nitrogen is 28

In 60 gm of urea nitrogen present = 28 gm

In 100 gm of urea nitrogen present\(= (C_3H_5O_2)_2 = C_6H_{10}O_4\)

Anhydrous \(CuSO_4\) is used to test presence of water in any liquid because it changes its colour white to blue.

Distillation particularly fractional because the boiling point of benzene (80°C) and chloroform (61.5°C) are close. Fractional distillation involves repeated distillations and condensations, in a fractionating column.

As a result of distillation and condensation at each point of the fractionating column, the vapours rising up become richer in more volatile component and the liquid falling back into the flask becomes richer in less volatile component.

Thus, the low boiling liquid distils first while the higher boiling liquid distils afterwards.

Halogens are detected by Beilstein’s test. In this test, a copper wire is dipped in original solution and heated in a bunsen burner flame. Green colour is imparted to the flame, due to the formation of a volatile copper halide. This proves the presence of halogen.

Liquid ammonia is used as a coolant in ice factories and cold storages.