Question 1

Question 9Define solubility. Give the main steps with the calculations involved of the method to determine the solubility of a given salt 'X' in water.

Accepted Answer

The solubility of a solute in a solvent at a particular temperature is the maximum amount of the solute in grams that will saturate 100 g. of the solvent at that temperature.Determination of Solubility of 'X' In Water :Step 1 : Preparation of saturated solution of 'X'Take a boiling tube half filled with about 100 ml of distilled water.Add crystals of 'X' to distilled water and stir slowly.Continue stirring till the crystals dissolve. Repeat the process till no more salt can dissolve.Pour the saturated solution — in a clean dry test-tube.Step 2 :(ii) Calculations To Determine The Solubility Of Solute :Weigh a clean and dry evaporating dish = M gAdd the above saturated solution to it and reweigh = M1 gHeat the solution to dryness and reweigh the dish with residue = M2g∴ Weight of solute = (M2 – M) gNote the temperature of saturated solution = t° CSolubility = Weight of soluteWeight of solvent\dfrac{	ext{Weight of solute}}{	ext{Weight of solvent}}Weight of solventWeight of solute​ x 100Solubility = [M2–M] g[M1−M2] g\dfrac{[	ext{M}_2 – 	ext{M]} 	ext{ g}}{	ext{[M}_1 - 	ext{M}_2] 	ext{ g} }[M1​−M2​] g[M2​–M] g​ x 100 at t° C

Question 2

Question 16Differentiate between hydrated and anhydrous crystals with examples. State three defined changes which occur when hydrated copper sulphate is heated.

Accepted Answer

Hydrated CrystalsAnhydrous SubstanceHydrated Crystals contain fixed number of water molecules [as water of crystallization] in loose chemical combination with the substance e.g., CuSO4.5H2OAnhydrous Substance does not contain any fixed number of molecules [as water of crystallization] in loose chemical combination with the substance e.g., KNO3Changes that occur when hydrated copper sulphate is heated are :White powder of anhydrous copper sulphate is formed from blue crystals of hydrated copper sulphate.No geometric shape is seen.Anhydrous copper sulphate is amorphous in nature.

Question 3

Question 20Conc. H2SO4 acts as a 'drying agent' & a 'dehydrating agent'. Explain and differentiate the words in italics.

Accepted Answer

Drying agentDehydrating agentThe substances which can readily absorb or remove moisture from other substances are known as Drying agent.The substances which remove chemically combined water or elements from compounds, due to their strong affinity for water are called Dehydrating agents.Conc. sulphuric acid [as a drying agent] - absorbs only moisture from the substance and makes it dry without changing it's composition.Conc. sulphuric acid [as dehydrating agent] - removes water molecule from the composition of substance and reacts chemically and produces a new substance with new properties.

Question 4

Question 26State the disadvantages of hardness in water.

Accepted Answer

Hard water is unfit for laundries, unsafe for drinking, not suitable for preparing solutions and for industrial uses since it forms a crusty boiler scale or fur in boilers.

Question 5

Question 27Temporary hardness in water can be removed by boiling. Give balanced equations to explain how, hardness in temporary hard water is removed by boiling.

Accepted Answer

Temporary hardness in water is due to the presence of calcium or magnesium bicarbonates.On boiling temporary hard water, calcium and magnesium bicarbonates in water decompose to give insoluble carbonates, carbon dioxide and water.The insoluble carbonates are filtered out, CO2 gas escapes and water is rendered soft.Ca(HCO3)2Calcium bicarbonate→BoilCaCO3↓Calcium carbonate+H2O+CO2[g]\underset{	ext{Calcium bicarbonate}}{	ext{Ca(HCO}_3)_2}\xrightarrow{	ext{Boil}} \underset{	ext{Calcium carbonate}}{	ext{CaCO}_3↓} + 	ext{H}_2	ext{O} + 	ext{CO}_2 [	ext{g}]Calcium bicarbonateCa(HCO3​)2​​Boil​Calcium carbonateCaCO3​↓​+H2​O+CO2​[g]Mg(HCO3)2Magnesium bicarbonate→BoilMgCO3↓Magnesium carbonate+H2O+CO2[g]\underset{	ext{Magnesium bicarbonate}}{	ext{Mg(HCO}_3)_2}\xrightarrow{	ext{Boil}} \underset{	ext{Magnesium carbonate}}{	ext{MgCO}_3↓} + 	ext{H}_2	ext{O} + 	ext{CO}_2 [	ext{g}]Magnesium bicarbonateMg(HCO3​)2​​Boil​Magnesium carbonateMgCO3​↓​+H2​O+CO2​[g]

Question 6

Question 28Both temporary & permanent hardness in water can be removed by addition of washing soda. Give balanced equations for the same.

Accepted Answer

Below are the balanced equations for removal of temporary & permanent hardness in water by addition of washing soda:Permanent Hard Water:Na2CO3Washing soda+CaSO4Calcium sulphate⟶CaCO3↓Calcium carbonate+Na2SO4\underset{	ext{Washing soda}}{	ext{Na}_2	ext{CO}_3} + \underset{	ext{Calcium sulphate}}{	ext{CaSO}_4} \longrightarrow \underset{	ext{Calcium carbonate}}{	ext{CaCO}_3↓} + 	ext{Na}_2	ext{SO}_4Washing sodaNa2​CO3​​+Calcium sulphateCaSO4​​⟶Calcium carbonateCaCO3​↓​+Na2​SO4​Temporary Hard Water:Na2CO3Washing soda+Ca(HCO3)2Calcium bicarbonate⟶CaCO3↓Calcium carbonate+2NaHCO3\underset{	ext{Washing soda}}{	ext{Na}_2	ext{CO}_3} + \underset{	ext{Calcium bicarbonate}}{	ext{Ca(HCO}_3)_2}\longrightarrow \underset{	ext{Calcium carbonate}}{	ext{CaCO}_3↓} + 2	ext{NaHCO}_3Washing sodaNa2​CO3​​+Calcium bicarbonateCa(HCO3​)2​​⟶Calcium carbonateCaCO3​↓​+2NaHCO3​

Question 7

Question 29A sample of water is given in a trough. State how would you prove experimentally whether the given sample is hard water or soft water.

Accepted Answer

A piece of ordinary soap is rubbed inside the sample of water. If the sample of water lathers, it is soft else it is hard.

Question 8

Question 30Two samples of water are placed in a beaker individually. State how you will determine experimentally which of the two samples contains permanent hard water.

Accepted Answer

Boil each sample of water separately. The boiled water is filtered. The clear residual sample left behind is rubbed with soap. If the sample of water does not lather after filtration, then the sample is permanent hard water.

Question 9

Question 31State what are synthetic detergents. Explain experimentally how you will determine the advantage of synthetic detergents over soap using a sample of hard water.

Accepted Answer

Synthetic detergents are sodium salts of sulphonic acid and their calcium and magnesium salts are soluble in hard water and do not form scum. Hence, synthetic detergents when rubbed in hard water lather easily and hence have an advantage over ordinary soap.In order to study the advantage of synthetic detergents over soap experimentally, a sample of water is divided into two parts.A synthetic soap is added to the first sample of hard water and rubbed slowly with the hand.An ordinary soap is then added to the second sample of water and rubbed similarly.Result : The sample of hard water lathers with a synthetic detergent but not with an ordinary soap.Reason for the observation:Ca(HCO3)2In hard water+2NaStSoap⟶Ca(St)2↓Scum+2NaHCO3\underset{	ext{In hard water}}{	ext{Ca(HCO}_3)_2} + \underset{	ext{Soap}}{2	ext{NaSt}} \longrightarrow \underset{	ext{Scum}}{	ext{Ca(St)}_2↓} + 2	ext{NaHCO}_3In hard waterCa(HCO3​)2​​+Soap2NaSt​⟶ScumCa(St)2​↓​+2NaHCO3​

Question 10

Question 12State the importance of the solubility of CO2, and O2 in water.

Accepted Answer

Solubility of CO2, and O2 in water is important for the following reasons:Marine life like fish use the oxygen of the air dissolved in water for respiration and thus aquatic life is sustained. 1 dm3 (1 litre) of water contains 40 cm3 of dissolved oxygen.Aquatic plants make use of dissolved carbon dioxide for photosynthesis, i.e., to prepare their food.6CO2 + 12H2O →sunlightChlorophyll\xrightarrow[	ext{sunlight}]{	ext{Chlorophyll}}Chlorophyllsunlight​ C6H12O6 + 6O2 + 6H2OCarbon dioxide dissolved in water reacts with limestone to form calcium bicarbonate.CaCO3 + CO2 + H2O ⟶ Ca(HCO3)2Marine organisms such as snails, oysters, etc., extract calcium carbonate from calcium bicarbonate to build their shells.

Question 11

Question 15Explain why:(a) Boiled or distilled water tastes flat.(b) Ice at zero degree centigrade has greater cooling effect than water at 0°C.(c) Burns caused by steam are more severe than burns caused by boiling water.(d) Rivers and lakes do not freeze easily?(e) Air dissolved in water contains a higher proportion of oxygen.(f) If distilled water is kept in a sealed bottle for a long time, it leaves etchings on the surface of the glass.(g) Rain water does not leave behind concentric rings when boiled.

Accepted Answer

(a) Pure water is tasteless. The taste in water is due to the gases and solids dissolved in it i.e., impurities present in it. As boiled and distilled water are pure containing no impurities hence they taste flat i.e., are tasteless.(b) The latent heat of fusion of ice is 336 J g-1 so it absorbs 336 J of heat and changes to water at 0°C and then water at 0°C absorbs heat and temperature is raised. Therefore, ice at 0°C absorbs extra heat in comparison to water at 0°C (due to latent heat of fusion). Hence, ice at 0°C has greater cooling effect than water at 0°C.(c) Steam has a higher heat content on account of high specific latent heat of condensation that is 2268 J g-1. Hence, steam at 100°C carries more heat than water. Therefore, burns caused by steam are more severe than burns caused by boiling water.(d) Rivers and lakes do not freeze easily because the specific latent heat of fusion of ice is sufficiently high (= 336 J g-1). The water in lakes and rivers will have to liberate a large quantity of heat to the surrounding before freezing. The layer of ice formed over the water surface, being a poor conductor of heat, will also prevent the loss of heat from the water of lake, hence the water does not freeze all at once.(e) Air dissolved in water contains a higher proportion of oxygen because oxygen is more soluble in water compared to nitrogen. The composition of air dissolved in water is 33% oxygen compared to 21% in ordinary air.(f) Substances that are apparently insoluble in water actually dissolve in it in traces. Even, when we put water in a glass vessel, an extremely small amount of glass dissolves in it. It is for this reason that when distilled water is kept in a sealed bottle for a long time, it leaves etchings on the inside surface of the glass.(g) As rainwater does not contain dissolved solids hence, it does not leave behind concentric rings when boiled.

Question 12

Question 5Give equations to show what happens when temporary hard water is(a) boiled(b) treated with slaked lime

Accepted Answer

(a) By boiling carbon dioxide is driven off and the soluble hydrogen carbonates are converted into insoluble carbonates and could be removed by filtration or decantation.Calcium carbonate and magnesium carbonate are precipitated leaving the water soft.Ca(HCO3)2 →Boil \xrightarrow{	ext{Boil}\space}Boil ​ CaCO3 ↓ + H2O + CO2 ↑Mg(HCO3)2 →Boil \xrightarrow{	ext{Boil}\space}Boil ​ MgCO3 ↓ + H2O + CO2 ↑(b) Lime stone is first thoroughly mixed with water in a tank and then fed into another tank containing the hard water. Revolving paddles thoroughly mix the two solutions.Most of calcium carbonate settles down. If there is any solid left over, it is removed by a filter. This process goes by the name 'Clark's process'.Ca(HCO3)2 + Ca(OH)2 ⟶ 2CaCO3 ↓ + 2H2OMg(HCO3)2 + Ca(OH)2 ⟶ MgCO3 ↓ + CaCO3 ↓ + 2H2O

Question 13

Question 8Explain with equation, what is noticed when permanent hard water is treated with(a) slaked lime(b) washing soda

Accepted Answer

(a) In order to treat permanent hard water, slaked lime is used, magnesium hydroxide and calcium carbonate precipitates out and can be easily filtered.MgSO4 + Na2CO3 + Ca(OH)2 ⟶ Mg(OH)2 ↓ + CaCO3 ↓ + Na2SO4(b) Washing soda is added to hard water, which results in settling down of insoluble carbonates which can be removed by filtration.Na2CO3Washing soda+CaSO4Calcium sulphate⟶CaCO3↓Calcium carbonate+Na2SO4\underset{	ext{Washing soda}}{	ext{Na}_2	ext{CO}_3} + \underset{	ext{Calcium sulphate}}{	ext{CaSO}_4} \longrightarrow \underset{	ext{Calcium carbonate}}{	ext{CaCO}_3↓} + 	ext{Na}_2	ext{SO}_4Washing sodaNa2​CO3​​+Calcium sulphateCaSO4​​⟶Calcium carbonateCaCO3​↓​+Na2​SO4​Na2CO3Washing soda+CaCl2Calcium chloride⟶CaCO3↓Calcium carbonate+2NaCl\underset{	ext{Washing soda}}{	ext{Na}_2	ext{CO}_3} + \underset{	ext{Calcium chloride}}{	ext{CaCl}_2} \longrightarrow \underset{	ext{Calcium carbonate}}{	ext{CaCO}_3↓} + 2	ext{NaCl}Washing sodaNa2​CO3​​+Calcium chlorideCaCl2​​⟶Calcium carbonateCaCO3​↓​+2NaCl

Question 14

Question 10Explain, with equations, why ordinary soap does not lather easily in hard water.

Accepted Answer

If the water is hard, calcium and magnesium ions of the water combine with the negative ions of the soap to form a slimy precipitate of insoluble calcium and magnesium usually called soap curd (scum).2NaSt[soap]+Ca(HCO3)2⟶CaSt2[soap curd]↓+2NaHCO3\underset{	ext{[soap]}}{2	ext{NaSt}} + {	ext{Ca(HCO}_3)_2} \longrightarrow \underset{	ext{[soap curd]}}{	ext{CaSt}_2} ↓ + {2	ext{NaHCO}_3}[soap]2NaSt​+Ca(HCO3​)2​⟶[soap curd]CaSt2​​↓+2NaHCO3​2NaSt+Mg(HCO3)2⟶MgSt2↓+2NaHCO3{2	ext{NaSt}} + {	ext{Mg(HCO}_3)_2} \longrightarrow {	ext{MgSt}_2} ↓ + {2	ext{NaHCO}_3}2NaSt+Mg(HCO3​)2​⟶MgSt2​↓+2NaHCO3​In permanent hard water, the formation of soap- curd will go on as long as there are calcium and magnesium ions present.2NaSt+CaCl2⟶CaSt2↓+2NaCl{2	ext{NaSt}} + {	ext{CaCl}_2} {\longrightarrow} {	ext{CaSt}_2} ↓ + {2	ext{NaCl}}2NaSt+CaCl2​⟶CaSt2​↓+2NaCl2NaSt+MgCl2⟶MgSt2↓+2NaCl{2	ext{NaSt}} + {	ext{MgCl}_2} {\longrightarrow} {	ext{MgSt}_2} ↓ + {2	ext{NaCl}}2NaSt+MgCl2​⟶MgSt2​↓+2NaCl2NaSt+CaSO4⟶CaSt2↓+2Na2SO4{2	ext{NaSt}} + {	ext{CaSO}_4} {\longrightarrow} {	ext{CaSt}_2} ↓ + {2	ext{Na}_2{	ext{SO}_4}}2NaSt+CaSO4​⟶CaSt2​↓+2Na2​SO4​2NaSt+MgSO4⟶MgSt2↓+2Na2SO4{2	ext{NaSt}} + {	ext{MgSO}_4} {\longrightarrow} {	ext{MgSt}_2} ↓ + {2	ext{Na}_2{	ext{SO}_4}}2NaSt+MgSO4​⟶MgSt2​↓+2Na2​SO4​As long as the formation of soap-curd continues, no soap lather will be formed and the cleaning of cloth or body will not be possible.

Question 15

Question 11Explain:(a) The use of lead pipes for drinking water supply is being discontinued.(b) Chalk hills dissolve in rain water.(c) Hard water is unfit for boilers.(d) Iron chloride forms a saturated solution when exposed to the atmosphere.(e) A bottle containing concentrated H2SO4 should be stoppered.

Accepted Answer

(a) Lead is toxic. When water flows through lead pipes, especially soft or acidic water, it can dissolve lead ions into the water. Ingesting lead causes serious health issues, especially in children (like brain and kidney damage).(b) Chalk is mainly calcium carbonate (CaCO3). Rainwater absorbs carbon dioxide (CO2) from the air and forms carbonic acid (H2CO3), a weak acid:CO2 + H2O ⟶ H2CO3This reacts with calcium carbonate:H2CO3 + CaCO3 ⟶ Ca(HCO3)2The product, calcium bicarbonate, is soluble in water, so the chalk slowly dissolves.(c) The dissolved substance present in the hard water does not convert into steam and gets deposited on the inner walls of the tube. Hence, the tubes become narrower and eventually less steam is produced.When bore of the tube becomes very narrow, the pressure of steam increases so much that at times the boiler itself bursts. Hence, hard water is unfit for use in boilers.(d) Ferric chloride (FeCl3) is deliquescent, it absorbs moisture from the air and dissolves in the water it absorbs, eventually forming a saturated solution. This property makes it unstable in open air.(e) Concentrated sulphuric acid H2SO4 is highly hygroscopic, meaning it absorbs water vapour from the air. If left open, it will absorb moisture and become diluted. To maintain concentration and prevent hazards, it should always be tightly stoppered.

Question 16

Question 1(iii)Permanent hardness of water is removed by:Adding calcium sulphateBoiling with potassium chlorideBoilingAdding sodium carbonate

Accepted Answer

Adding sodium carbonateReason — Permanent hardness of water can be removed by addition of washing soda [sodium carbonate]:Permanent Hard Water:Na2CO3Washing soda+CaSO4Calcium sulphate⟶CaCO3↓Calcium carbonate+Na2SO4\underset{	ext{Washing soda}}{	ext{Na}_2	ext{CO}_3} + \underset{	ext{Calcium sulphate}}{	ext{CaSO}_4} \longrightarrow \underset{	ext{Calcium carbonate}}{	ext{CaCO}_3↓} + 	ext{Na}_2	ext{SO}_4Washing sodaNa2​CO3​​+Calcium sulphateCaSO4​​⟶Calcium carbonateCaCO3​↓​+Na2​SO4​

Question 17

Question 1(xiii)Temporary hardness of water can be removed by:Adding sodium chlorideBoilingAdding calcium carbonateLeaving it for a few hours.

Accepted Answer

BoilingReason — By boiling carbon dioxide is driven off and the soluble hydrogen carbonates are converted into insoluble carbonates and could be removed by filtration or decantation.Calcium carbonate and magnesium carbonate are precipitated leaving the water soft.Ca(HCO3)2 →Boil \xrightarrow{	ext{Boil}\space}Boil ​ CaCO3 ↓ + H2O + CO2 ↑Mg(HCO3)2 →Boil \xrightarrow{	ext{Boil}\space}Boil ​ MgCO3 ↓ + H2O + CO2 ↑

Question 18

Question 1(xv)Sodium chloride (common salt) besides being used in kitchen can also be used as the raw material for making:P — Slaked limeQ — Washing sodaR — Baking sodaOnly POnly QOnly RBoth Q and R

Accepted Answer

Both Q and RReason — Washing Soda (NaHCO3.10H2O) and baking Soda (NaHCO3) are made from sodium chloride using the Solvay processNaCl + NH3 + CO2 + H2O ⟶ NaHCO3 + NH4ClNaHCO3 is filtered and heated to get sodium carbonate: ​2NaHCO3 →Δ\xrightarrow{\Delta}Δ​ Na2CO3 + CO2 + H2OSodium carbonate (Na2CO3) is then hydrated to form washing soda.

Question 19

Question 2(ii)Assertion (A): A saturated solution becomes unsaturated on heating.Reason (R): More amount of solute can dissolve in a solvent upon heating.Both A and R are true and R is the correct explanation of A.Both A and R are true but R is not the correct explanation of A.A is true but R is false.A is false but R is true.

Accepted Answer

Both A and R are true and R is the correct explanation of A.Explanation — When a saturated solution is heated to a higher temperature, then it becomes unsaturated. More solute can be dissolved in this solution now. Hence, both the assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question 20

Question 2(iii)Assertion (A): A white powder forms on the surface of washing soda crystals which are left exposed to the air.Reason (R): Washing soda is a hygroscopic substance.Both A and R are true and R is the correct explanation of A.Both A and R are true but R is not the correct explanation of A.A is true but R is false.A is false but R is true.

Accepted Answer

A is true but R is false.Explanation — Washing soda (hydrated sodium carbonate), when exposed to dry air, loses water of crystallization. The result is the formation of white anhydrous sodium carbonate (Na2CO3) on the surface, which appears as a white powder. Hence the assertion (A) is true. Washing soda undergoes efflorescence, meaning it loses moisture (its water of crystallization) to the air. So washing soda is efflorescent, not hygroscopic. Hence the reason (R) is false.

Water

Additional Questions

Exercise 3A

Exercise 3C Descriptive Type

Exercise 3C Multiple Choice Type

Exercise 3C Short Answer Type

Problems Based On Solubility

Unit Test Paper 3 — Water