M K Verma Solutions for Chapter: Purification of Organic Compounds, Exercise 1: TOPICWISE QUESTIONS

Chromatography is a valuable method for the separation, isolation, purification and identification of the constituents of a mixture and it is based on the general principle of:

$50 \mathrm{mg}$ of an organic compound containing nitrogen was analyzed by Kjeldahl's method and the evolved ammonia was absorbed in $20 \mathrm{ml}$ of $0.1 \mathrm{M}$ ${\mathrm{H}}_2{\mathrm{SO}}_4$solution. The excess of acid required $25 \mathrm{ml} \text{of} 0.1 \mathrm{M} \mathrm{NaOH}$ solution for complete neutralization. The percentage of nitrogen in the compound is

$0.24 \mathrm{g}$ of an organic compound gives $0.22 \mathrm{g} {\mathrm{CO}}_2$ on complete combustion. If it contains $16.6\%$ hydrogen, the percentage of carbon and oxygen, respectively are:

Molecular weight of an organic compound can be determined by:

An organic compound $(\mathrm{mole} \mathrm{weight}=60)$ is found to contain $\mathrm{C}=20\%, \mathrm{H}=6.67\% \mathrm{and} \mathrm{N}=46.67\%$, while the rest is oxygen. On heating, it gives ${\mathrm{NH}}_3$ along with a solid residue. The solid residue gives a violet colour with alkaline ${\mathrm{CuSO}}_4$. The compound is:

$0.24 \mathrm{g}$ of a volatile liquid on vaporization gives $45 \mathrm{mL}$ of vapour at NTP. What will be the vapour density of the substance? (Density of ${\mathrm{H}}_2 = 0.089 \mathrm{g} {\mathrm{L}}^{-1}$)

$0.759 \mathrm{g}$ of a silver salt of a monobasic organic acid on the ignition left $0.463 \mathrm{g}$ metallic silver. The equivalent weight of the acid is:

Given: Atomic mass of silver =$108$.

$0.32 \mathrm{g}$ of an organic compound in the Carius method gave $0.233 \mathrm{g} \text{of} {\mathrm{BaSO}}_4$. The percentage of sulphur in the organic compound is:

Given the atomic mass of sulphur and barium, respectively: $\left(32 \mathrm{and} 233\right)$.