Chemistry

This is a long answer type question as classified in NCERT Exemplar

(i)Mass percentage $\frac{\mathbf{w}}{\mathbf{w}}$ : It is a weight by weight relationship. Thus mathematically ,

Mass % of a component = $\mathrm{}\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}}{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}$

where , W represents weight of the component or solution

(ii) Volume percentage $\frac{\mathbf{V}}{\mathbf{V}}$ :

Volume % of a component= $\mathrm{}\frac{\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}}{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}$

where , V represents volume

(iii) It is mass by volume percentage $\frac{\mathbf{w}}{\mathbf{V}}$ : Thus mathematically, where W is weight of solute in grams , but measures volume of solution in milliliters (ml).

$\frac{\mathbf{w}}{\mathbf{V}}$ = $\mathrm{}\frac{\mathrm{w}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}}{\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}$

(iv) Parts per million or ppm is a unit of measure of dissolved solids in solution, in terms of a ratio between the number of parts of solids/solute to a million parts  of total volume. Thus, 1 Parts per million (ppm):- is a concentration of solution that contains 1 gram solute and 1000000 ml solution (same as 1 milligram solute per litre of solution) 

(v) Mole fraction of a component (x)= $\frac{\mathrm{N}\mathrm{o}.\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}}{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}}$

(vi) The no. of moles of solute per litre of solution.

Molarity (M )= $\frac{\left(\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\right)}{\left(\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\right)}$

(vii) The no. of moles of solute per kg of solvent.

Molality (m) = $\frac{\left(\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}\right)}{\left(\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{k}\mathrm{g}\right)}$

Effect of temperature: mass %, ppm, mole fraction  and molality do not change with temperature whereas molarity, volume percentage and mass by volume percentage changes with temperature because volume of solution (liquid) varies with changes in temperature.

This is a  Short Answers Type Questions as classified in NCERT Exemplar

Scientists working in the Antarctica region reported that depletion of the ozone layer  is commonly termed as an ozone hole over the South Pole in the Antarctic region. It was observed that a unique set of conditions were responsible for the ozone hole. In summer, nitrogen dioxide and methane react with chlorine monoxide and chlorine atoms forming chlorine sinks, preventing much ozone depletion, whereas in winter, special types of clouds called polar stratospheric clouds are formed over Antarctica. These polar stratospheric clouds provide a surface on which chlorine nitrate gets hydrolysed to form hypochlorous acid. It also reacts with hydrogen chloride produced to give molecular chlorine.

When sunlight returns to the Antarctica in the spring, then the sun's warmth breaks up the clouds and HOCl and Cl2are photolyzed in the presence of sunlight. The chlorine radicals thus formed, initiate the chain reaction for ozone depletion. and the reactions can be given as-

HOCl (g) OH (g) + Cl (g)

Cl2 (g) 2Cl (g)