AP 9th Class Physical Science Chapter 4 Solution – Is Matter Pure
Andhra Pradesh SCERT 9th Class Physical Science Chapter 4 Is Matter Pure Solution for AP 9th Class Physics/Chemistry Exam. Lots of Students of Andhra Pradesh Board will search on internet for Andhra Pradesh Class 9 Physical Science Textbook Solution or Study Material for AP 9th exam. Here you search will end! Here in this page we have provided for all question answer for AP SCERT Chapter 4 Is Matter Pure.
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AP 9th Class Chemistry Chapter 4 Is Matter Pure Solution
1.) Which separation techniques will you apply for the separation of the following?(AS1)
(a) Sodium chloride from its solution in water.
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
(c) Small pieces of metal in the engine oil of a car.
(d) Different pigments from an extract of flower petals.
(e) Butter from curd.
(f) Oil from water.
(g) Tea leaves from tea.
(h) Iron pins from sand.
(i) Wheat grains from husk.
(j) Fine mud particles suspended in water.
Ans:- a) Evaporation ; b) Sublimation ; c) Filtration ; d) Chromatography ; e) Centrifugation ; f) Using separating funnel ; g) Filtration ; h) Using magnet ; i) Winnowing ; j) Centrifugation.
2.) Write the steps you would use for making tea. Use the words given below and writethe steps for making tea? (AS7)
Solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Ans: –The steps for making tea is –
- Take 100ml water as a solvent.
- After adding 1 teaspoon of sugar and tea leaves boiled the mixture to dissolve this solutes.
- After that add some water as a solute and boil it.
- Then after few minutes filtering the solution and the tea leaves which are present at filter is residue.
3.) Explain the following giving examples. (AS1)
(a) Saturated solution (b) Pure substance (c) colloid (d) Suspension
Ans: – a) There is no possibility to dissolve any solute without applying different temperature, these solution is called saturated solution.
b) If a substance made up with the only one type of particle is known as pure substance.
c) In a heterogeneous mixture where tyndall effect shows and the size of the particles in between the size of the particle of suspension and solutions is known as colloid.
d) The heterogeneous mixture in whichthe particles are visible in naked eyes as these are not dissolve in the mixture, this types of solution is known as suspension.
4.) Classify each of the following as a homogeneous or heterogeneous mixture. Give reasons.(AS1)
Soda water, wood, air, soil, vinegar, filtered tea.
Ans: – There is uniform composition and appearance in a homogeneous mixture. And the different phases are visible in a heterogeneous mixture.
Homogeneous mixture – soda water,air,vinegar,filtered tea.
Heterogeneous mixture – soil, wood.
5.) How would you confirm that a colourless liquid given to you is pure water? (AS1)
Ans: – we can confirm the purity of a colourless liquid by boiling. At 100℃ temperature the liquid show boiling then it is pure.
6.) Which of the following materials fall in the category of a “pure substance”? Givereasons (AS1)
(a) Ice (b) Milk (c) Iron (d) Hydrochloric acid (e) Calcium oxide
(f) Mercury (g) Brick (h) Wood (i) Air.
Ans: – In a pure substance there is no impurity present in it.
The material which are pure is ice, iron, hydrochloric acid,calcium oxide and mercury.
7.) Identify the solutions among the following mixtures. (AS1)
(a) Soil (b) Sea water (c) Air (d) Coal (e) Soda water.
Ans: – The mixture among them are sea water, air, soda water.
8.) Which of the following will show “Tyndall effect”? How can you demonstrate Tyndall effect in them? (AS1 AS3)
(a) Salt solution (b) Milk (c) Copper sulphate solution (d) Starch solution.
Ans: – As we all know that the colloidal solution show the tyndall effect. From these example only milk and starch are colloids so these two only show the tyndall effect.
9.) Classify the following into elements, compounds and mixtures.(AS1)
(a) Sodium (b) Soil (c) Sugar solution (d) Silver
(e) Calcium carbonate (f) Tin (g) Silicon (h) Coal
(i) Air (j) Soap (k) Methane (l) Carbon dioxide
(m) Blood
Ans: –Element: – sodium, silver, tin and silicon.
Compounds: -calcium carbonate, methane and carbon dioxide.
Mixture: -soil, sugar solution, coal, air, soap and blood.
10.) Classify the following substances in the below given table. (AS1)
Ink, soda water, brass, fog, blood, aerosol sprays, fruit salad, black coffee, oil and water, boot polish, air, nail polish, starch solution, milk.
Ans: –Solution: -Ink, black coffee, water, soda, brass.
Suspension: -fruit salad, oil and water.
Emulsion: – blood, boot polish, nail polish, milk.
11.) Take a solution, a suspension, a colloidal dispersion in different beakers. Test whether each of these mixtures shows the Tyndall effect by focusing a light at the side of the container. (AS3)
Ans: –Let take solution of sugar, a suspension as chalk powder with water and milk like colloidal solution. And separate them in three different bekar.
After stir the three bekar pass the light through them. Then we will see that the light passes through the second and third bekar.
After some time again when we passes light through them we will see light only passes through the third bekar where milk are kept.
By this experiment we came to this conclusion that as milk act as colloidal solution it only shows the tyndall effect.
12.) Draw the figures of arrangement of apparatus for distillation and fractional distillation. What do you find the major difference in these apparatus? (AS5)
Ans:-
13.) Determine the mass by mass percentage concentration of a 100g salt solution which contains 20g salt? (AS1)
Ans: – From question it is given that, mass of the salt (solute) =20g and mass of solution is 100g.
So mass by mass percentage = (mass of solute / mass of solution) ×100 = 20/100×100= 20%.
14.) Calculate the concentration interms of mass by volume percentage of the solution containing. 2.5g potassium chloride in 50ml of potassium chloride (KCl) solution? (AS1)
Ans: – As we all know,
Mass by volume of any solution is = (mass of the solute/ volume of the solution) ×100.
From question it given that, mass of potassium (solute) =2.5g,and volume of potassium chloride (solution volume) = 50ml.
Putting this value in the equation we get,2.5/50×100 = 5%.