11th Standard Syllabus & Materials
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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Chemistry Test1.
In the estimation of nitrogen present in an organic compound by Dumas method 0.35 g yielded 20.7 mL of nitrogen at 150 C and 760 mm pressure. Calculate the percentage of nitrogen in the compound.
2.
0.32 g of an organic compound, after heating with fuming nitric acid and barium nitrate crystals is a sealed tube game 0.466 g of barium sulphate. Determine the percentage of sulphur in the compound.
3.
0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water calculate the percentage of carbon and hydrogen in it.
4.
Explain paper chromatography.
5.
Describe optical isomerism with suitable example.
1.
\( \mathrm{w} = 0.35 \mathrm{~g} \)
\(\mathrm{~V}_1 = 31.7 \mathrm{~mL}=31.7 \times 10^{-3} \mathrm{~L} \)
\(\mathrm{~T}_1 = 15^{\circ} \mathrm{C}=273+15=288 \mathrm{k} \)
\(\mathrm{P}_1 = 760-12.8=747.2 \mathrm{~atm} \)
\(\mathrm{~V}_0 = ? \)
\(\mathrm{P}_0 = 760 \mathrm{~atm} \mathrm{~T}_0 = 273 \mathrm{~K} \)
\( \frac{\mathrm{P}_0 \mathrm{~V}_0}{\mathrm{~T}_0}=\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1} \text { (or) } \mathrm{V}_0=\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1} \times \frac{\mathrm{T}_0}{\mathrm{P}_0} \)
\(\mathrm{~V}_0=\frac{747.2 \times 31.7 \times 10^{-3} \times 273}{288 \times 760}=2.95 \times 10^{-2} \mathrm{~L} \)
\(\therefore \% \mathrm{~N}=\ \frac{28}{22.4} \times \frac{\mathrm{V}_0}{\mathrm{w}} \times 100\)
2.
w = 0.32 g and x = 0.466 g
\(
\% \mathrm{~S} =\frac{32}{233} \times \frac{\mathrm{x}}{\mathrm{w}} \times 100
\)
\(=\frac{32}{233} \times \frac{0.466}{0.32} \times 100=20 \%\)
3.
W = 0.3 g
x = 0.54 g
y = 0.88 g
\(
\% \mathrm{C} =\frac{12}{44} \times \frac{\mathrm{y}}{\mathrm{w}} \times 100=\frac{12}{44} \times \frac{0.88}{0.30} \times 100=80 \%
\)
\(\% \mathrm{H}=\quad \frac{2}{18} \times \frac{\mathrm{x}}{\mathrm{w}} \times 100=\frac{2}{18} \times \frac{0.54}{0.30} \times 100=20 \%\)
4.
Partition Chromatography:
Paper chromatography (PC) is an example of partition chromatography. The same procedure is followed as in thin layer chromatography except that a strip of paper acts as an adsorbent is method involves continues differential portioning of components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as chromatography paper is used. This paper act as a stationary phase.
A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which act as the mobile phase solvent rises up and flows over the spot. The paper selectively retains different components according to their different partition in the two phases where a chromatogram is developed. The spots of the separated colored compounds are visible at different heights from the position of initial spots on the chromatogram. The spots of the separated colorless compounds may be observed either under ultraviolent light or by the use of an appropriate spray reagent.
5.
Optical isomerism:
Compounds having same physical and chemical property but differ only in the rotation of plane of the polarized light are known as optical isomers and the phenomenon is known as optical isomerism.
Some organic compounds such as glucose have the ability to rotate the plane of the plane polarized light and they are said to be optically active compounds and this property of a compound is called optical activity. The optical isomer, which rotates the plane of
the plane polarised light to the right or in cloclauisl direction is said to be dextrorotary (dexter means right) denoted by the sign (+), whereas the compound which rotates to the left or anticloclanrise is said to be leavo rotatory (leanues mean left) denoted by sign(-).
Dextrorotatory compounds are represented as 'd' or by sign (+) and lavorotatory compounds are ( - ) represented as 'l' or by sign (-).
Enantiomerism and optical activity: An optically active substance may exist in two or more isomeric forms which have same physical and chemical properties but differ in terms of direction of rotation of plane polarized light, such optical isomers which rotate the plane of polarized light with equal angle but in opposite direction are known as enantiomers and the phenomenon is knovrrn as enantiomerism. Isomers which are non-super impossible mirror images of each other are called enantiomers.
conditions for enantiomerism or optical isomerism:
A carbon atom whose tetra vaiency is satisfied by four different substituents (atoms or groups) is called a symmetric carbon or chiral carbon. It is indicated by an asterisk as C*. A molecule Possessing chiral carbon atom and non-super impossible to its own mirror image is said to be a chiral molecule or asymmetric, and the pioperty is called chirality or dissymmetry.

11th Standard Syllabus & Materials
11th Standard
TN 11th Tamil பீடு பெற நில் - செய்யுள் - காவடிச்சிந்து Important Questions And Answers Study Material - QB365 Set A
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