CHAPTER-2
IS MATTER AROUND US PURE
Exercise
Q1- 1. Which separation techniques will you apply for the separation of the following?
(a) Sodium chloride from its solution in water.
Evaporation
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
Sublimation
(c) Small pieces of metal in the engine oil of a car.
Filtration or Centrifugation or decantation
(d) Different pigments from an extract of flower petals.
Chromatography
(e) Butter from curd.
Centrifugation
(f) Oil from water.
Using separating funnel
(g) Tea leaves from tea.
Filtration
(h) Iron pins from sand.
Magnetic separation
(i) Wheat grains from husk.
Winnowing
(j) Fine mud particles suspended in water.
Centrifugation
Q2- Write the steps you would use for making tea. Use the words: solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Answer
First, water is taken as a solvent in a saucer pan. This water (solvent) is allowed to boil. During heating, milk and tea leaves are added to the solvent as solutes. They form a solution. Then, the solution is poured through a strainer. The insoluble part of the solution remains on the strainer asresidue. Sugar added to the filtrate, which dissolves in the filtrate. The resulting solution is the required tea.
Q3- . Pragya tested the solubility of three different substances at different temperatures and collected the data as given below( results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).
(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?
(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.
(c) Find the solubility of each salt at 293 K. What salt has the highest solubility at this temperature?
(d) What is the effect of change of temperature on the solubility of a salt?
Answer
(a) Since 62 g of potassium nitrate is dissolved in 100g of water to prepare a saturated solution at 313 K, 31 g of potassium nitrate should be dissolved in 50 g of water to prepare a saturated solution at 313 K.
(b) The amount of potassium chloride that should be dissolved in water to make a saturated solution increases with temperature. Thus, as the solution cools some of the potassium chloride will precipitate out of the solution.
(c) The solubility of the salts at 293 K are:
Potassium nitrate – 32 g
Sodium chloride – 36 g
Potassium chloride – 35 g
Ammonium chloride – 37 g
Potassium nitrate – 32 g
Sodium chloride – 36 g
Potassium chloride – 35 g
Ammonium chloride – 37 g
Ammonium chloride has the highest solubility at 293 K.
(d) The solubility of a salt increases with temperature.
Q4- . Explain the following giving examples:
(a) Saturated solution
(b) Pure substance
(c) Colloid
(d) Suspension
Answer
(a) Solution in which no more solute can be dissolved at a particular temperature is known as saturated solution. For example in aqueous solution of sugar no more sugar can be dissolved at room temperature.
(b) A pure substance is a substance consisting of a single type of particles i.e., all constituent particles of the substance have the same chemical properties. For example water, sugar, salt etc.
(c) A colloid is a heterogeneous mixture whose particles are not as small as solution but they are so small that cannot be seen by naked eye. When a beam of light is passed through a colloid then the path of the light becomes visible. For example milk, smoke etc.
(d) A suspension is a heterogeneous mixture in which solids are dispersed in liquids. The solute particles in suspension do not dissolve but remain suspended throughout the medium. For example Paints, Muddy water chalk water mixtures etc.
Q5- . Classify each of the following as a homogeneous or heterogeneous mixture.
Soda water, wood, air, soil, vinegar, filtered tea
Answer
Homogeneous mixtures: Soda water, air, vinegar, filtered tea
Heterogeneous mixtures: Wood, soil
Note: Pure air is homogeneous mixture but Polluted air is heterogeneous mixture.
Q6- . How would you confirm that a colourless liquid given to you is pure water?
Answer
Take a sample of colourless liquid and put on stove if it starts boiling exactly at 100 ºC then it is pure water. Any other colourless liquid such as vinegar always have different boiling point. Also observe carefully that after some time whole liquid will convert into vapour without leaving any residue.
Q7- . Which of the following materials fall in the category of a "pure substance"?
(a) Ice
(b) Milk
(c) Iron
(d) Hydrochloric Acid
(e) Calcium oxide
(f) Mercury
(g) Brick
(h) Wood
(i) Air
Answer
The following materials fall in the category of a "pure substance":
(a) Ice
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury
Q8- . Identify the solutions among the following mixtures:
(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water
Answer
The following mixtures are solutions:
(b) Sea water
(c) Air
(e) Soda water
Q9- . Which of the following will show the "Tyndall effect"?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution
Answer
Tyndall effect is shown by colloidal solution. Here milk and starch solution are colloids therefore milk and starch solution will show Tyndall effect.
Q10-. Classify the following into elements, compounds and mixtures:
(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
Answer
Elements: Sodium, Silver, Tin and Silicon.
Compounds: Calcium carbonate, Methane and carbon dioxide.
Mixtures: Soil, Sugar, Coal, Air, Soap and Blood.
Q11- . Which of the following are chemical changes?
(a) Growth of a plant
(b) Rusting of iron
(c) Mixing of iron fillings and sand
(d) Cooking of food
(e) Digestion of food
(f) Freezing of water
(g) Burning of candle
Answer
The following changes are chemical changes:
(a) Growth of a plant
(b) Rusting of iron
(d) Cooking of food
(e) Digestion of food
The following changes are chemical changes:
(a) Growth of a plant
(b) Rusting of iron
(d) Cooking of food
(e) Digestion of food
(g) Burning of candle
Ans given
Q a train started from rest to attain a velocity of 72km /h in 5 min. find acceleration and distance?
V=72km/h
· Crop variety improvement
· Higher yield
Soil supplies rest of nutrient.
Manure is classified into compost and vermicompost.
2. How is a prokaryotic cell different from a eukaryotic cell?
Answer
3. What would happen if the plasma membrane ruptures or breaks down?
Answer
If the plasma membrane ruptures or breakdown then the cell will not be able to exchange material from its surrounding by diffusion or osmosis. Thereafter the protoplasmic material will be disappeared and the cell will die.
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4. What would happen to the life of a cell if there was no Golgi apparatus?
Answer
Golgi apparatus has the function of storage modification and packaging of the products. If there is no Golgi apparatus then the packaging and transporting of materials synthesized by cell will not happen.
5. Which organelle is known as the powerhouse of the cell? Why?
Answer
Mitochondria are known as the powerhouse of cells because energy required for various chemical activities needed for life is released by mitochondria in the form of ATP (Adenosine triphosphate) molecules.
6. Where do the lipids and proteins constituting the cell membrane get synthesized?
Answer
Lipids are synthesized in Smooth endoplasmic reticulum (SER) and the proteins are synthesized in rough endoplasmic reticulum (RER).
7. How does an Amoeba obtain its food?
Answer
Amoeba takes in food using temporary finger-like extensions of the cell surface which fuse over the food particle forming a food-vacuole as shown in figure. Inside the food vacuole, complex substances are broken down into simpler ones which then diffuse into the cytoplasm. The remaining undigested material is moved to the surface of the cell and thrown out.
8. What is osmosis?
Answer
Osmosis is the process in which water molecules moves from the region of high concentration to a region of low concentration through a semi permeable membrane.
9. Carry out the following osmosis experiment:
Take four peeled potato halves and scoop each one out to make potato cups. One of these potato cups should be made from a boiled potato. Put each potato cup in a trough containing water. Now,
(a) Keep cup A empty
(b) Put one teaspoon sugar in cup B
(c) Put one teaspoon salt in cup C
(d) Put one teaspoon sugar in the boiled potato cup D.
Keep these for two hours. Then observe the four potato cups and answer the following:
(i) Explain why water gathers in the hollowed portion of B and C.
(ii) Why is potato A necessary for this experiment?
(iii) Explain why water does not gather in the hollowed out portions of A and D.
Answer
(i) Water gathers in the hollowed portions of set-up B and C because water enters the potato as a result of osmosis. Since the medium surrounding the cell has a higher water concentration than the cell, the water moves inside by osmosis. Hence, water gathers in the hollowed portions of the potato cup.
(ii) Potato A in the experiment acts as a control set-up. No water gathers in the hollowed portions of potato A.
(iii) Water does not gather in the hollowed portions of potato A because potato cup A is empty. It is a control set-up in the experiment.
Water is not able to enter potato D because the potato used here is boiled. Boiling denatures the proteins present in the cell membrane and thus, disrupts the cell membrane. For osmosis, a semi-permeable membrane is required, which is disrupted in this case. Therefore, osmosis will not occur. Hence, water does not enter the boiled potato cup.
Motion
when an object move from one place to another place with the respect of time :
Is called motion.
Distance :
- it is the total path covered by a moving object.
- It is denoted by ‘s’ and its unit is metre.
- It is scalar quantity.
Speed:
- speed is the distance covered by an object per unit time .
- Speed is denoted by ‘v’
- s.i unit of speed is m/s
- it is scalar quantity. And speed =distance /time
Displacement :
- it is the shortest distance between the final point to intinal point .
- it is the vector quantity .
- s.i unit of displacement =m/s.
velocity :
- it is defined as displacement per unit time.
- velocity is denoted by “V”
- velocity is a vector quantity
- s.i unit of velocity is m/s
acceleration :
- it is defined as the rate of change of velocity.
- Acceleration =v-u/t
- Si unit of acceleration is m/s2
Retardation:
- retardation is the change in velocity per unit time when the intinal velocity is more than the final velocity.
- s.i unit is m/s2
equation of motion
- v=u+at
- s=ut+1/2at2
- v2=u2-2as
Example
Q; the break applied to a car produce acceleration of 6ms2in the opposite direction to the motion .
if the car take 2s tostop after the application of brake . calculate the distance during this time.
Ans given
A=6m/s2
V=o,t= 2s
According to the first equation of motion
V=u+at
O=u+6*2 ,u=12
Acccortion to the second equation of the motion
S=utr+at2
=24-12
=12m
Q : a car accelerate uniformly 18km/h to 36km/h in 5s calculate acceleration and distance?
Ans u=18km/h, v=36km/h ,t= 5s
From first eqation of motionV=u+at10=5+5a10-5=5aA= 5/5According to the 2nd equation of motionS=ut+at2 =25+1/2*25 25+12.5=37.5m
Q a train started from rest to attain a velocity of 72km /h in 5 min. find acceleration and distance?
Ans u=om/s
V=72km/h
T= 5miin
According to the firs t equation of motion
V=u+at
20=0+300a
300a=20
A=20/300
1/15
According to the 2nd equation of motion.
S=ut+1/2at2
=o*300+1/2*1/5*300
3000m
3km
6. A car accelerates uniformly from 18 km h–1 to 36 km h–1 in 5 s. Calculate (i) the acceleration and
ANS- U= 18 km h-1 = 5 m s-1 , V= 36 km h-1 = 10 m s-1 , t = 5 s
ANS- a = -6 m s-2 , t = 2 sec , V= 0 m s-2
6. A car accelerates uniformly from 18 km h–1 to 36 km h–1 in 5 s. Calculate (i) the acceleration and
(ii) the distance covered by the car in that time.
ANS- U= 18 km h-1 = 5 m s-1 , V= 36 km h-1 = 10 m s-1 , t = 5 s
(i) a = V-U = 10-5 = 1 m s-2
t 5
(ii) by II nd equation of motion
S= ut + 1/2 at*2
S= 5X5 + 1/2 X 1 X 25
S= 25 + 12.5 = 37.5 m
7. The brakes applied to a car produce an acceleration of 6 m s-2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.
ANS- a = -6 m s-2 , t = 2 sec , V= 0 m s-2
by I st equation of motion
V=U+at
0= U+(-12)
0=U-12
U= 12 m s-1
by II nd equation of motion
S= ut + 1/2 at*2
S= 12X2 + 1/2 X -6 X 4
S= 24 - 12 = 12 m
DISTANCE TIME GRAPH
Improvement in food
Resources
1. Plant2. Animal
Production of plant that we obtain as food
· Cereals=give carbohydrates
· Pulse=give protein
· Fruit and vegetable = give mineral and vitamin
The activity which involved in improving crop production
· Crop variety improvement
· Crop production improvement
· Crop protection improvement
Crop variety improvement
· Higher yield
· Improvement in quality
· Biotic and abiotic resistance
· Change in maturity duration=the duration of crop should be shorter . so that gap between the sowing the harvest is less which make the crop yield economic
· Wider adaptability=the crop should be raised in such a manner that a plant variety can adapt to the environment and climatic condition.
Crop production management
· Nutrient management=nutrient are supplied to plant through air water and soil .there are 16nutrient which are essential to plant .air supply carbon and oxygen . water supplies hydrogen.
Soil supplies rest of nutrient.
Out 13 nutrient 6 are require in large amount is known as macron nutrient.and 7 are require in small quantity is micron nutrient.
· Manure=it is prepared by decomposition of animal excetra and plant waste.
Manure is classified into compost and vermicompost.
Compost=compost is formed in the field by diging a pit and filling it with animal waste and plant .
Vermicompost=the compost is prepared by using red worm to hasten the process of decomposition.
Green manure =in the field the plant are grown like sun hemp
And gaur and the field are ploughed in it.these plant mix with the soil and converted into manure.
· Fertilizer=are commercially produced nutrient specific chemical which are produced in factory.they supplied potassium nitrogen and phaphosrous.
· Irrigation=irrigation is the process of watering plant.
1. Dug well=the water is collect from the lower starta of the earth.
2. Tube well=the water is collect from the deeper starta of te earth .
3. Canal=it take water from the river and are divded into many branch.
4. Water lift system=in this system the water is directly lifted from the river .
5. Tanks=are small reservore which are the catchment of run of water.
· Crop pattern=different type of the crop are grown in the same piece of land to get the maximum benefit.
1. Mixed crop=different type of crop grown in the same piece of land.
2. Inter cropping=it is the growing of different in definite pattern.
EXERCISE
1. Who discovered cells and how?
Answer
An English Botanist, Robert Hooke discovered cells. In 1665, he used self-designed microscope to observe cells in a cork slice.
2. Why is the cell called the structural and functional unit of life?
Answer
Cells are called the structural and functional unit of life because all the living organisms are made up of cells and also all the functions taking place inside the body of organisms are performed by cells.
Answer
An English Botanist, Robert Hooke discovered cells. In 1665, he used self-designed microscope to observe cells in a cork slice.
2. Why is the cell called the structural and functional unit of life?
Answer
Cells are called the structural and functional unit of life because all the living organisms are made up of cells and also all the functions taking place inside the body of organisms are performed by cells.
1. Make a comparison and write down ways in which plant cells are different from animal cells.
Answer
Answer
| Animal cell | Plant cell |
| The do not have cell wall. | They have cell wall made up of cellulose. |
| They do not have chloroplast. | They contain chloroplast. |
| They have centrosome. | They do not have centrosome. |
| Vacuoles are smaller in size. | Vacuoles are larger in size. |
| Lysosomes are larger in number. | Lysosomes are absent or very few in number |
| Prominent Golgi bodies are present. | Subunits of Golgi bodies known as dictyosomes are present. |
2. How is a prokaryotic cell different from a eukaryotic cell?
Answer
| Prokaryotic cell | Eukaryotic cell |
| Most prokaryotes are unicellular. | Most eukaryotes are multicellular. |
| Size of the cell is generally small (0.5- 5 µm). | Size of the cell is generally large (50- 100 µm). |
| Nuclear region is poorly defined due to the absence of a nuclear membrane or the cell lacks true nucleus. | Nuclear region is well-defined and is surrounded by a nuclear membrane, or true nucleus bound by a nuclear membrane is present in the cell. |
| It contains a single chromosome. | It contains more than one chromosome. |
Nucleolus is absent.
| Nucleolus is present. |
| Membrane-bound cell organelles such as plastids, mitochondria, endoplasmic reticulum, Golgi apparatus, etc. are absent. | Cell organelles such as mitochondria, plastids, endoplasmic reticulum, Golgi apparatus, lysosomes, etc. are present. |
| Cell division occurs through binary fission | Cell division occurs by mitosis. |
| Prokaryotic cells are found in bacteria and blue-green algae. | Eukaryotic cells are found in fungi, plants, and animal cells. |
3. What would happen if the plasma membrane ruptures or breaks down?
Answer
If the plasma membrane ruptures or breakdown then the cell will not be able to exchange material from its surrounding by diffusion or osmosis. Thereafter the protoplasmic material will be disappeared and the cell will die.
Page No: 67
4. What would happen to the life of a cell if there was no Golgi apparatus?
Answer
Golgi apparatus has the function of storage modification and packaging of the products. If there is no Golgi apparatus then the packaging and transporting of materials synthesized by cell will not happen.
5. Which organelle is known as the powerhouse of the cell? Why?
Answer
Mitochondria are known as the powerhouse of cells because energy required for various chemical activities needed for life is released by mitochondria in the form of ATP (Adenosine triphosphate) molecules.
6. Where do the lipids and proteins constituting the cell membrane get synthesized?
Answer
Lipids are synthesized in Smooth endoplasmic reticulum (SER) and the proteins are synthesized in rough endoplasmic reticulum (RER).
7. How does an Amoeba obtain its food?
Answer
Amoeba takes in food using temporary finger-like extensions of the cell surface which fuse over the food particle forming a food-vacuole as shown in figure. Inside the food vacuole, complex substances are broken down into simpler ones which then diffuse into the cytoplasm. The remaining undigested material is moved to the surface of the cell and thrown out.
Answer
Osmosis is the process in which water molecules moves from the region of high concentration to a region of low concentration through a semi permeable membrane.
9. Carry out the following osmosis experiment:
Take four peeled potato halves and scoop each one out to make potato cups. One of these potato cups should be made from a boiled potato. Put each potato cup in a trough containing water. Now,
(a) Keep cup A empty
(b) Put one teaspoon sugar in cup B
(c) Put one teaspoon salt in cup C
(d) Put one teaspoon sugar in the boiled potato cup D.
Keep these for two hours. Then observe the four potato cups and answer the following:
(i) Explain why water gathers in the hollowed portion of B and C.
(ii) Why is potato A necessary for this experiment?
(iii) Explain why water does not gather in the hollowed out portions of A and D.
Answer
(i) Water gathers in the hollowed portions of set-up B and C because water enters the potato as a result of osmosis. Since the medium surrounding the cell has a higher water concentration than the cell, the water moves inside by osmosis. Hence, water gathers in the hollowed portions of the potato cup.
(ii) Potato A in the experiment acts as a control set-up. No water gathers in the hollowed portions of potato A.
(iii) Water does not gather in the hollowed portions of potato A because potato cup A is empty. It is a control set-up in the experiment.
Water is not able to enter potato D because the potato used here is boiled. Boiling denatures the proteins present in the cell membrane and thus, disrupts the cell membrane. For osmosis, a semi-permeable membrane is required, which is disrupted in this case. Therefore, osmosis will not occur. Hence, water does not enter the boiled potato cup.
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