11th Class Chemistry Chapter 9 Solutions Short Questions Answers
11th Class Chemistry Chapter 9 Solutions Short Questions Answers
We are providing all Students from 5th class to master level all exams preparation in free of cost. Now we have another milestone achieved, providing all School level students to the point and exam oriented preparation question answers for all science and arts students.
After Online tests for all subjects now it’s time to prepare the next level for Punjab board students to prepare their short question section here. We have a complete collection of all classes subject wise and chapter wise thousands questions with the right answer for easy understanding.
Here we are providing complete chapter wise Chemistry questions and Answers for the 11th class students. All the visitors can prepare for their 11th class examination by attempting below given question answers.
In this List we have included all Punjab boards and both Arts and Science students. These Boards students can prepare their exam easily with these short question answer section
Lahore Board 11th classes short questions Answer
Rawalpindi Board 11th classes short questions Answer
Gujranwala Board 11th classes short questions Answer
Multan Board 11th classes short questions Answer
Sargodha Board 11th classes short questions Answer
Faisalabad Board 11th classes short questions Answer
Sahiwal Board 11th classes short questions Answer
DG Khan Board 11th classes short questions Answer
Bahwalpur Board 11th classes short questions Answer
All above mention Punjab Boards students prepare their annual and classes test from these online test and Short question answer series. In coming days we have many other plans to provide all kinds of other preparation on our Gotest website.
How to Prepare Punjab Board Classes Short Question Answer at Gotest
- Just Open the desired Class and subject which you want to prepare.
- You have Green bars which are Questions of that subject Chapter. Just click on Bar, it slides down and you can get the right answer to those questions.
- You can also Rate those question Answers with Helpful or not to make it more accurate. We will review all answers very carefully and also update time to time.
Now you can start your preparation here below
[toggle title=”1. How will you prepare 5% w/v urea solution in water?” state=”close”]We will dissolve 5 grams of urea and make the total volume of solution as 100 cm3. In this solution, we know the mass of solute and not the mass or volume of solution.[/toggle]
[toggle title=”2. In weight %, we know the masses of solute and solvent but in weight % age, we don’t know the mass of solvent and that of the solution. Give comments?” state=”close”]w/w % is weight of the solute dissolved per 100 parts by weight of solution. In 5% w/w glucose solution. We know the weight of the solute. Wt. of the solvent and wt. of the solution. In w/v glucose solution, we dissolves 5 g of glucose is 100 cm3 of solution. So, we do not know the final mass of solvent and the weight of solution.[/toggle]
[toggle title=”3. The concentration in terms of modality is independent of temperature but morality depends upon the temperature. How?” state=”close”]In molal solutions the mass of the solvent and that of the solute are also fixed. The mass of the substances are not temperature dependent. In molar solutions we have the volumes of solutions. Volume of a liquid is a temperature independent. So the morality is not influenced by temperature but molarity does change.[/toggle]
[toggle title=”4. The sum of the mole fractions of all the components is always equal to unity for any solution?” state=”close”]Mole fraction is the ratio of number of moles of one component to total number of moles. Hence mole fraction of any substance is always less than unity. The mole fraction of three components A, B and C in a solutions are
XA = nA XB = nB/nA + nB + nC, Xc = nc/ nA + nB + nC
nA+nB + nc ,
Adding these mole fractions:
XA + XB + Xc = nA/nA + nB+ nC + nB/ nA + nB + nC + nC / nA + nB + nc = 1[/toggle]
[toggle title=”5. What are mathematical formulas of molarity and molality?” state=”close”]
Molarity is the number of moles of the solute per dm3 of the solution. It has to the following formul:
Molarity (M) = Mass of solute 1
Molar mass of solute * Volume of solute in dm3
Molarity (M) = Number of moles of solu
Volume of solution in dm3
Molality (M)= Mass of solute
Molar mass of solute * Mass of solvent in kg
[/toggle]
[toggle title=”6. 100 g of 98 % H2SO4 has a volume of 54.34 cm3 of H2SO4 because its density is 1.84 g cm-3. Justify it?” state=”close”]
98% H2SO4 means that 100 g of H2SO4 solution has 98 g of H2 So4 and only 2 g of water.
Density = mass /volume
Since, volume = mass/ density
Putting the values
Volume = 100 g = 54.34cm3
1.84 g cm3
It means that the 98% H2SO4 having total mass of 100 g has a volume of 54.34.
[/toggle]
[toggle title=”7. The total volume of solution by mixing 100 cm3 of water with 100 cm3 of alcohol may not be equal to 200 cm3. It is true?” state=”close”]Molecules of water and alcohol have bent structures. The forces of attractions among water and alcohol in pure states are greater than after mixing. So the total volume of the mixture of the two components is greater than the sum of individual volumes.[/toggle]
[toggle title=”8. One molal of solution of urea in water in dilute as compared to one molar solution of urea but the number of particles of the solute is same. Justify it?” state=”close”]In one molal solution of urea, 60 g of urea is dissolved in 100 g of water, which is approximately 1000cm3 of water. In one molar solution of urea, 60 g of urea is added in water to make total volume of solution as 1000cm3. So the volume of water in molar solution is less than 1000cm3. Hence, molar solution is concentrated and molal solution is dilute.[/toggle]
[toggle title=”9. Why glucose is not soluble in CCl4, but dissolves in water?” state=”close”]The molecule of glucose has strong hydrogen bonding due to the presence of five ― OH groups. CCl4 being non- polar molecules cannot break the hydrogen bonding among the glucose molecules. Water is a polar solvent, so its breaks the hydrogen bonding of glucose and dissolves it. Similarly can sugar is not soluble in benzene.[/toggle]
[toggle title=”10. CaCl2. 2H2O shows discontinuous solubility curve, when plotted against temperature. Why?” state=”close”]The changes of temperature change the existent of hydrogen. The solubility change with temperature depends upon the number of water molecules in the substance.[/toggle]
[toggle title=”11. What do you mean by discontinuous solubility curves?” state=”close”]The graph between solubilities and temperature may show a sudden change in the direction. Such a graph is called discontinuous solubility curve. Such curves are the combinations of two or more than two curves due to the changing behavior of that compound.[/toggle]
[toggle title=”12. How does fractional crystallization help in removing the impurities from a solid substance?” state=”close”]The impure substance is dissolved is hot solvent. The solute to be purified should be less soluble than impurities. When cooling is done, this solute settles down in the form of crystals and impurities are left behind in the solution. [/toggle]
[toggle title=”13. What is conjugate solution?” state=”close”]When two partially miscible liquids are mixed, two layers are produced. They are the solutions of one component into the other. These two solutions have distinct line of demarcation are the called conjugate solutions.[/toggle]
[toggle title=”14. What is effect of temperature on the conjugate solution of water and phenol?” state=”close”]Water and phenol are partially miscible. They make to distinct solutions and two separate layers at room temperature. By increasing the temperature, the composition of the two conjugate solution change and at 65.90C a homogeneous mixture of the two components is formed. [/toggle]
[toggle title=”15. What is consulate temperature or critical solution temperature?” state=”close”]partially miscible liquids make two distinct layers of conjugate solutions. By changing the temperature, the two layers become completely miscible and homogeneous solution is produced. This temperature is called consulate temperature. [/toggle]
[toggle title=”16. How does an increase in temperature may increase or decrease the solubility of a substance?” state=”close”]When the heat of solution ΔHv is positive and such solutions are heated, they dissolve more solutes and the solubility increases. When the heat of solution is negative, then the heat is being evolved during the formation of solution. When such solutions are heated, their solubility’s.Decrease.[/toggle]
[toggle title=”17. Give two substance of Raoult’s law?” state=”close”](i). The vapor pressure o solvent in the solution states is directly proportional the mole fraction of solvent p = p0 * X1 (ii). Lowing of V.P. is directly portional to be mole fraction of solute ΔP = P0 X2.[/toggle]
[toggle title=”18. The boiling point of one molal urea solution is 100.520C, but the boiling point of two molal urea solution is less than 101. 040C. Why?” state=”close”]Two molal urea solution is a concentrated solution as compared to one molal urea solution. The number of molecules of urea in a two molal solution are twice the Avogradro’s number but the number of particles which are free to move in the solution are less than twice the Avogadro’s number. Because one dm3 volume is less to accommodate 2NA molecules of solute and behave as non- ideal solution to show colligative property. So the elevation of B.P is not doubled.[/toggle]
[toggle title=”19. Why the relative lowering of vapour pressure is independent of temperature?” state=”close”]The relative lowering of vapour pressure and mole fraction of solute are related as:
Δp / P0 = X2
Vapour pressure and lowering of vaour pressure upon temperature. So, when the temperature of a solution is increased both the factors ΔP and P0 increase in such a way that the ratio remains the same.[/toggle]
[toggle title=”20. When we plot a graph between temperature and compositions of binary liquid mixture, straight line is not obtained. We get two curves even for ideal solutions. Justify it?” state=”close”]Different solutions of two components have their own percentage in the liquid states and the vapour state. So, if the graph is plotted between compositions on Y – axis, than we will get two curves. The lower curve is for the % of liquid mixture at various temperatures and the upper curve is for % of vapours of various mixtures.[/toggle]
[toggle title=”21. Non- ideal solutions do not obey the Raoult’s law. Why?” state=”close”]The molecules of components in non- ideal solutions have forces of attractions for each other. The values of vapour pressures of individuals’ components are not proportional to their mole fractions as in Raoult’s law. [/toggle]
[toggle title=”22. What is negative deviation from Raoult’s law?” state=”close”]The pairs of liquids which have lesser values of the vapour pressure for some of their composition than those of pure components show negative deviation from Raoult’s law. They show a minimum point in the vapour pressure composition curve.[/toggle]
[toggle title=”23. What is positive deviation from Raoult’s law?” state=”close”]The pairs liquids which have together value of vapour pressure for some of their composition than those of pure components show positive deviation from Raoult’s law. They show a maximum point in the vapour pressure composition curve.[/toggle]
[toggle title=”24. What do you mean by minimum boiling point mixture?” state=”close”]This is an azeotropic mixture which shows positive deviation from Raoult’s law. It boils at low temperature than either of pure components.[/toggle]
[toggle title=”25. Define colligative properties? Name some important colligative properties?” state=”close”]Those properties of solutions which depend upon the number of the particles of the solute and not upon their nature are called colligative properties. These properties are (i), Lowering of vapour pressure (ii), Elevation of boiling point (iii). Depression of freezing point (iv). Osmotic pressure [/toggle]
[toggle title=”26. Colligatave properties are obeyed when the solute is non- electrolyte and also when the solutions are dilute?” state=”close”]In the case of electrolytes, ions are produced, number of particles of the solutes increase and the amounts of colligative properties also increase. Colligative properties are obeyed when the solutions are dilute, so that the solute particles are bevaving independently.[/toggle]
[toggle title=”27. Boiling points of solvents increase due to the presence of solutes. Why?” state=”close”]The surface of the solution has molecules of solute as well. They do not allow the solvent to leave the surface as rapidly as in pure solvent. To boil the solutions, we have to increase the temperature of solutions in comparison to pure solvents. So, the B.P of solutions are higher the pure solvents. [/toggle]
[toggle title=”29. Why the freezing points are depressed due presence of solutes?” state=”close”]The lowering of vapour pressure completes the solutions to freeze at those temperatures, which are below the freezing point of pure solvent. The reason is that the vapour pressure temperature curve meets the solids phase of pure solvent at lower temperature than the pure solvent.[/toggle]
[toggle title=”30. What is ebullioscopic constant?” state=”close”]It is elevation of the boiling point of the solvent when one mole of the non- volatile, non- electrolyte solute is dissolved is 1 Kg of the solvent. It is property of solvent and not the solute. It is denoted by Kb. Its value is 0.520 for water.[/toggle]
[toggle title=”31. What is cryoscopic constant?” state=”close”]It is depression of freezing point of the solvent, when one mole of non- volatile, non- electrolyte is dissolved in 1 kg of the solvent. It is property of the solvent not of the solute. It is denoted by Kf. Its value is 1.860C for water.[/toggle]
[toggle title=”32. Why Beckmann thermometer is used to not the depression of freezing point?” state=”close”]Beckmann thermometer can measure up to 1/20th of the degree. The elevation of boiling points and the depression of freezing points for dilute solutions are very small quantities. Hence, one can measure these very small changes of temperatures. [/toggle]
[toggle title=”33. The lowering of vapour pressure, elevation of boiling point and this depression of freezing points are call collligative properties. Comment upon it?” state=”close”]Collagative properties are those properties which depend upon the number of particles of the solute in the solution. If we dissolve equal number of moles of non- volatile and non- electrolyte solutions in the same quantity of the solvent, then the lowering of vapour pressure, elevation of boiling point and the depression of freezing point will be the same because the number of particles are same. [/toggle]
[toggle title=”34. Why a non- volatile solute in a volatile solvent lowers the vapour pressure of solution?” state=”close”]The particles of the solute are distributed throughout the bulk of the solution and some of the particles of the solute are also present on the surface of the solution. The number of molecules of the solvent per unit area on the surface of the solution because less. Hence the evaporating tendency of the solvent decreases and vapour pressure of solution becomes less.[/toggle]
[toggle title=”35. How the relative lowering of vapour pressure can help us to calculate the molar mass of a non- volatile, non- electrolyte solute?” state=”close”]According to the third definition of the Raoult’s law, the relative lowering of vapour pressure is equal to the mole fraction of the solute.
Δp/p0 = X2
Since X2 = n2/n1 + n2 = n2/n1 (n1>> n2)
So, Δp/p0 = W2/M2 * M1/W1
or M2 = W2/W1 . M1 . p0/Δp
From these equation molar mass can be calculated.[/toggle]
[toggle title=”36. Why the boiling point of a solution of a non- volatile solute in a volatile solvent is always greater than the boiling point of a pour solvent?” state=”close”]The elevation of the boiling point of the solution is due to the lowering of vapour pressure of the solution as compared to the pour solvent. In the case of solution vapour pressure becomes less. In order to equalize the vepour pressure of the solution to the external pressure, more temperature is required than that in the pure state of the solvent, which causes elevation of boiling point. [/toggle]
[toggle title=”37. Why the freezing point of the solution is always less than the freezing point of the pour solvent?” state=”close”]The depression of the freezing point of the solution is again due lowering of vapour pressure of the solution as compared to the pour solvent. A graph between temperature and vapor pressure shows that the curve for the touches the curve for the solid phase of the pour solvent at a lower temperature, than the freezing point of the pour solvent.[/toggle]
[toggle title=”38. Prove that the collgative properties like ΔTb and ΔTf are inversely proportional to the molar masses of the solute?” state=”close”]
[/toggle]
[toggle title=”39. Colligative properties are obeyed when solutions are dilute. Why?” state=”close”]Colligative properties depend upon the number of particles of the solute. In dilute solutions the particles of the solute are for away and independent from each other and their number is according to the concept of molarity. When the solutions are concentrated, the two or more than two particles of the solute may remain together and may not behave independently. So, concentrated solution shows abnormal colligative properties.[/toggle]
[toggle title=”40. If an electrolyte is dissolved in a solvent, the abnormal colligative properties are observed. Why?” state=”close”]Electrolyte are those substances which dissociates into positive and negative ions. In this way the number of particles dissociated in a solution increase. The increased number of particles increases the colligative properties, called abnormal colligative property.[/toggle]
[toggle title=”41. In summer the antifreeze solutions protect the radiator from boiling over, why?” state=”close”]Water boils at 1000C. It is used in the radiators to decrease the temperature of the working engine. If we add some suitable solutes which increase the boiling point of water, above 1000C, then easy boiling over of water is avoided. Actually such solutes also decrease the F.P of solutions as well.[/toggle]
[toggle title=”42. Why the NaCl and KNO3 are used to lower the melting point of ice?” state=”close”]NaCl and KNO3 are electrolyte and are sufficiency soluble in water. They double the number of particles after dissociated in water. In this way they, can manage to decrease the freezing point of water to a greater extent as compared to a non- electrolyte. [/toggle]
[toggle title=”43. When heat of solution is negative, then increase in temperature decreases the solubility and vice versa. Why?” state=”close”]It is amount heat evolved when one mole of a substance is dissolved in excess of solvent. When the heat of solution is negative, it means that the vessel is heated up during the solution formation. So, when heat is supplied from outside, then the system will go to that side where greater amount of heat can be stored and that is the side of low solubility. [/toggle]
[toggle title=”44. Heat of solution of a substance is measured at infinite dilution. Why?” state=”close”]Actually the maximum amount of heat is evolved or absorbed, when each particle of one mole of solute is completely solvated by the solvent molecules. These solvated particles of the solute should be far away from each other in the solution. This is only possible when this solution is very dilute.[/toggle]
[toggle title=”45. How the ions are stabilized when a strong electrolyte like NaCl and KCl are dissolved in H2O?” state=”close”]Strong electrolyte is dissociated to the maximum extent and positive and negative ions are produced. These ions are surrounded by the water molecules as follows. Their charges are satisfied by the solvent molecules. In this way, they get the stabilized. [/toggle]
[toggle title=”46. How the forces of attractions between the ions and the solvent molecules depend upon the charge densities of the ions?” state=”close”]Smaller the size of the ion or greater the amount of the positive change of the ion, greater the charge density. Such ions have greater forces of attractions for the molecules of the solvent. They are solvated more effectively. Negatively charged ions are bigger are sizes, so they are not solvated in a better way.[/toggle]
[toggle title=”47. What are hydrates? How they are formed? Give some examples.” state=”close”]The crystalline substance which contains chemically combined water in definite proportion is called a hydrate. They are produced when aqueous solution of soluble salt is evaporated, Examples are
MgCl2. 6H2O. AlCl3. 6H2O, Na2CO3, 10H2O, CuSO4, 5H2O[/toggle]
[toggle title=”48. Why is salt produced from a strong acid and a strong base is not hydrolysed in water?” state=”close”]The positively charged change ion like Na+, K+, etc, are derived from strong bases are dissociated almost 100% in water. These positive ions have least tendencies to react with H2O molecules. The negatively charged ions like Cl– , Br–, 1–SO-24 etc, are derived from strong acids. These acids are dissociated almost 100% in water. The negative ions have little tendencies to react with H2O. In other words, they are not hydrolysed by h2O.[/toggle]
[toggle title=”49. Why a salt produced from a weak acid and a strong base gives a basic aqueous solution like CH3COONa?” state=”close”]The salt like CH3COO–N+a is formed from a strong base and a weak acid. It dissociated in water to give CH3COO– and Na+. Na+ does not react with water. CH3 COO– reacts with water to give CH3COOH and OH–. Due to free OH– ion the solution is basic.[/toggle]
[toggle title=”50. Explain why CuSO4 and NH4Cl give acid solution when put separately in water?” state=”close”]CuSO4 is a salt made up of a weak base, Cu(OH)2 and strong acid H2SO4. CuSO4 ionizes in water to give Cu+2 and SO4-2 ions. Copper ions are hydrolysed and H+ ions are set free in the solution. So solution is acidic. Similarly NH4Cl is made up of NH4Cl a weak base, and HCl a strong scid. Hence, Its aqueous solution is acidic. [/toggle]
Have Any other short question related to this 11th Class Chemistry Chapter 9 Solutions please write down in comment section.
