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Calculate the mass of sodium sulphate required to prepare its 20% (mass percent) solution in 100 g of water?

Solution:
Mass Percent: amount of solute in 100 gram of solution.
Mass Percent (%) =\frac{massofsolute\times100}{massofSolution}
Let the mass of sodium sulphate required be = m gram
Mass of the solvent (water) = 100 gram
The mass of solution = (m + 100) gram (sum of mass of solute and solvent)
20=\frac{m\times100}{m+100}
\Rightarrow 20m+2000=100m
\Rightarrow 2000=80m
\Rightarrow m=25g

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Arun has prepared 0.01\% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions?
(a) 1.00 g \;of\; NaCl + 100g \;of\; water

(b) 0.11 g \;of \;NaCl + 100 g\; of \;water

(c) 0.01 g \;of\; NaCl + 99.99 g \;of\; water

(d) 0.10 g \;of \;NaCl + 99.90 g \;of\; water

Answer: C
Solution:
Mass Percent: amount of solute in 100 gram of solution.
Mass percent (%)  =\frac{massofsolute\times100}{massofSolution}
Option a:
Mass percent (%) =\frac{1gram\times100}{1gram+100gram}=0.99
Option b:
Mass percent (%) =\frac{0.11gram\times100}{0.11gram+100gram}=0.1099
Option c:
Mass percent (%)  =\frac{0.1gram\times100}{0.1gram+99.99gram}=0.01
Option d:
Mass percent (%)  =\frac{0.1gram\times100}{0.1gram+99.90gram}=0.1
Hence the correct answer is C.

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You are provided with a mixture containing sand, iron filings, ammonium chloride and sodium chloride. Describe the procedures you would use to separate these constituents from the mixture?

Solution:
Given Mixture – Sand + Iron filings + Ammonium chloride +Sodium chloride
Iron filing can be separated by using magnet which will attract all the iron filings.
Now by the process of sublimation we can separate ammonium chloride from the mixture as it is the only constituent which get sublimate.
Now we mix the remaining salt and sand in water. The sodium chloride will dissolve in water and we can collect the sand which will settle in the base.
Now we are left with salt solution. By the process of evaporation water will vaporise and we are left with salt .

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During an experiment the students were asked to prepare a 10\% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution
(a) Are the two solutions of the same concentration?
(b) Compare the mass \% of the two solutions.

Solution:
Mass Percent: amount of solute in 100 gram of solution.
Mass percent(%) =\frac{massof Solute\times100}{massofSolution}
Ramesh’s Solution:
Mass percent(%) =\frac{10gram\times100}{10gram+100gram}=9.0909\%
Sarika’s Solution:
Mass percent(%) =\frac{10gram\times100}{100gram}=10\%
Part a: Sarika has higher mass percentage than Ramesh.
Part b: The mass percentage ratio of Sarika and Ramesh is 10: 9.0909

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Classify each of the following, as a physical or a chemical change. Give reasons.
(a) Drying of a shirt in the sun.
(b) Rising of hot air over a radiator.
(c) Burning of kerosene in a lantern.
(d) Change in the colour of black tea on adding lemon juice to it.
(e) Churning of milk cream to get butter.

Solution:
If a matter goes under a physical change its appearance might change but the substance of matter does not change.
In a chemical change in a chemical change either one of the substance changes and comes with new properties.
a) Drying of a shirt in the sun \RightarrowSubstance of shirt does not change\RightarrowPhysical change
(b) Rising of hot air over a radiator\RightarrowSubstance in air does not change\RightarrowPhysical change
(c) Burning of kerosene in a lantern\Rightarrow Kerosene goes under chemical reaction with oxygen\RightarrowChemical change
(d) Change in the colour of black tea on adding lemon juice to it\RightarrowBlack tea goes under chemical reaction\RightarrowChemical change
(e) Churning of milk cream to get butter\RightarrowSubstance in milk does not change\RightarrowPhysical change
(a, b, e) : Physical changes because there is no change in chemical composition, (c), (d) : Chemical changes because new substances are formed.

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A group of students took an old shoe box and covered it with a black paper from all sides. They fixed a source of light (a torch) at one end of the box by making a hole in it and made another hole on the other side to view the light. They placed a milk sample contained in a beaker/tumbler in the box as shown in the Fig. 2.4. They were amazed to see that milk taken in the tumbler was illuminated. They tried the same activity by taking a salt solution but found that light simply passed through it?

(a) Explain why the milk sample was illuminated. Name the phenomenon involved.
(b) Same results were not observed with a salt solution. Explain.
(c) Can you suggest two more solutions which would show the same effect as shown by the milk solution?

Scattering of light through heterogeneous mixture is called Tyndall effect.
Examples of Tyndall effect:
(i) In Fog, Beam of head light of car is visible.
(ii) Sunlight passing through dust particle will be visible.
(iii) Path of light rays seen in front of the projector in a cinema hall.
The Tyndall effect is shown by mixture, where mixture has tendency to scatter the light of shorter wavelength and transmit the light of larger wavelength.
The Tyndall effect is shown by heterogeneous mixture such as colloid.
(a) Milk is a colloidal solution hence shows Tyndall effect.
(b) Salt solution is homogeneous therefore do not show Tyndall effect and  they do not scatter light.
(c) Detergent mixture and air with dust will show Tyndall effect.

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A child wanted to separate the mixture of dyes constituting a sample of ink. He marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in Fig. 2.3. The filter paper was removed when the water moved near the top of the filter paper.


(i) What would you expect to see, if the ink contains three different coloured components?
(ii) Name the technique used by the child.
(iii) Suggest one more application of this technique.

Chromatography is a method, which is used to separate components of any mixture.
In this process, different component travels at different speeds when dissolved in solution.
Due to different speeds, all the components get separated from each other
(i) The components of the ink will travel with water at different speeds and we will see three bands on the filter paper at different levels.
(ii) Chromatography.
(iii) This technique is also used for separating pigments present in chlorophyll.

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Iron filings and sulphur were mixed together and divided into two parts, ‘A’ and ‘B’. Part ‘A’ was heated strongly while Part ‘B’ was not heated. Dilute hydrochloric acid was added to both the Parts and evolution of gas was seen in both the cases. How will you identify the gases evolved?

Solution:
Part A –
When iron and sulphur are heated they react with each other and form iron-sulphide.
Fe+S\rightarrow FeS
This iron sulphide can react with acid Hydrochloride and hydrogen Sulphide gas will come out from the mixture.
FeS+2HCl\rightarrow FeCl_{2}+H_{2}S(gas)
Hydrogen Sulphide has smell like rotten egg and can be identified easily.
This gas can be chemically tested by reaction with lead Salts.
Part B –
When iron and sulphur are heated they react with each other and form iron-sulphide but without heat they will remain unaffected.
Sulphur does not react with hydrochloride but iron will react and hydrogen gas will come out.
Fe+2HCl\rightarrow FeCl_{2}H_{2}(gas)
Hydrogen gas is highly inflammable and produces sound on burning.
If we bring any spark near the mixture, we can identify the hydrogen gas.

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(a) Under which category of mixtures will you classify alloys and why?
(b) A solution is always a liquid. Comment.
(c) Can a solution be heterogeneous?

 

Solution:
Any Homogeneous mixture of two substances is called Solution.
The substance in large quantity is called solvent and substance in small quantity is called Solute.
Solution can be gaseous (For example; Air), Liquid (For example; Salt solution with water) and Solid (For example; Alloy
(a) Alloys they have uniform composition throughout hence they are homogeneous solid solution.
(b) No, Any Solution can be gaseous (For example; Air), Liquid (For example; Salt solution with water) and Solid (For example; Alloy)
 (c) No, a solution is a homogeneous mixture and does not show Tyndall Effect.

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Fractional distillation is suitable for separation of miscible liquids with a boiling point difference of about 25 K or less. What part of fractional distillation apparatus makes it efficient and possess an advantage over a simple distillation process? Explain using a diagram.

If we mix two liquids of significant different boiling points, we can separate them by process of distillation.
Generally, this technique is used if difference in boiling point is more than 25^{\circ}C \sim 30^{\circ}C.
But Fractional Distillation is effective even the boiling point difference is less than 25^{\circ}C
In fractional distillation Method, a fractionating column is used and it makes it efficient. This column is packed with glass beads or small plates. It increases the surface area. The increase in surface area will increase the rate of condensation for the vapours. Vapour quickly loose energy, when they come in contact with beads or plates and can be condensed easily.
The length of the column would increase the efficiency of the process.

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