Chemistry Important Questions: Solutions and Colligative Properties (1 Mark)

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Section A – Questions of 1 mark VSA of difficulty level

Q1. What is the most important condition required for solute and solvent, while preparing a binary solution?

  • The condition required for solute and solvent, while preparing a binary solution is that the molecules of the different constituents of the solution (solute, solvent) must be so similar to one another that the molecules of one of the components can replace those of the other without variation of the spatial structure of the solution or of the energy due to intermolecular interactions.

Q2. Name the physical states of solute and solvent in the solution of iodine in CCl4 at room temperature.

  • In the solution of iodine in CCl4 at room temperature, the solute is iodine which is in the solid state while the solvent is CClwhich is in the liquid state.

Q3. Which is the best and flawless way of expressing the concentration of a solution?

  • Molality is a better way of expressing concentration than molarity because there is no term of volume of solvent is involved. The volume of the solvent depends on the temperature of the solvent. Thus there is no effect of the change of temperature on the molality.

Q4. Write reason, ‘concentration of a solution in terms of molality is preferred in comparison with molarity’.



  • Molality is a better way of expressing concentration than molarity because there is no term of volume of solvent is involved. The volume of the solvent depends on the temperature of the solvent. Thus there is no effect of the change of temperature on the molality.

Q.5. Which is a scientific and useful way of expressing the concentration of components of a solution?

  • Molality is a better way of expressing concentration than molarity because there is no term of volume of solvent is involved. The volume of the solvent depends on the temperature of the solvent. Thus there is no effect of the change of temperature on the molality.

Q6. Prove that “0.5 M ethyl alcohol and 0.1 M magnesium phosphate have same osmotic pressure at 25°C.”

  • Ethyl alcohol (C2H5OH) is non-electrolyte. It does not dissociate in water. Hence the concentration of solute particles, in this case, becomes approximately 0.5 M.
  • Magnesium phosphate Mg3(PO4)2 is an electrolyte. It dissociates as

Mg3(PO4)2  \rightleftharpoons   3Mg2+   + 2 (PO4)3-

  • Thus on dissociation, there are 5 solute particles. Hence the concentration of solute particles, in this case, becomes approximately 0. 1 x 5 = 0.5 M.
  • A freezing point depression is a colligative property which depends on the number of solute particles in the solution.
  • The number of solute particles in both the cases is the same. Hence their osmotic pressure is equal.

Q7. What is the most important condition, required for separation of solvent and solution in osmosis?



  • There is a difference in concentration of the solution and the solvent and the solution and solvent are separated by a semipermeable membrane. A semipermeable membrane is a membrane through which solvent particles can pass but the solute particles cannot.

Q8. Give a reason, “oxygen is slightly soluble in water; but highly soluble in blood.”

  • oxygen is slightly soluble in water but highly soluble in blood. this is because the vital constituent of blood, the haemoglobin react with oxygen as

Haemoglobin + 4O2 → Haemoglobin O8

  • Therefore the solubility of oxygen in blood increases. The oxygenated blood is pumped to various parts of the body for the supply of oxygen.

Q9. A person added a table salt in his carbonated soft drink and he was astonished. Write the reason on the basis of Henry’s law.

  • To increase the solubility of CO2 in carbonated soft drinks and soda water, the bottle is sealed under high pressure.
  • As table salt is added to carbonated soft drinks. The salt dissolves and the solubility of CO2 in water decreases and the carbon dioxide comes out of carbonated soft drinks in form of bubbles and soft drink oozes out of the bottle with effervescence.

Q10. If a non-volatile electrolyte produces four ions after dissociation then what will be the change in its colligative property?

  • Dissociation: Non-volatile electrolyte like inorganic acids, bases, and salts in aqueous solutions undergo dissociation, that is, the molecules break down into positively and negatively charged ions. In such cases, the number of effective particles increases.
  • Hence osmotic pressure, elevation of boiling point, and depression of freezing point are much higher than those calculated on the basis of an undissociated single molecule.

Q11. What is the most important condition required for any solution to be an ideal or nonideal?



  • The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions.
  • When a solution does not obey Raoult’s law over the entire range of concentration, then it is called a non-ideal solution.

Q12. Write the mathematical formula for the molar mass of non-volatile substance in cryoscopy.

  • Cryoscopy is the process of determining the lowered freezing points produced in a liquid by dissolved substances to determine molecular weights of solutes.
  • The mathematical formula for the molar mass of non-volatile substance in cryoscopy is

Solution

Where M2 = Molecular mass of solid

M1 = Molecular mass of solvent

W1 = Mass of solvent

W2 = Mass of solute



Kf = Freezing point depression constant or cryoscopic Constant.

ΔTf = Depression of freezing point

Q13. Give a reason, “in cold countries, the ice frozen on roads and footpaths, is melted by spraying salts like sodium chloride or calcium chloride.”

  • When some soluble salt like sodium chloride or calcium chloride is added to water, it decreases the freezing point of water, The phenomenon is called the depression in freezing point.
  • When salt is sprinkled over ice, it’s freezing point is lowered up to -20°C. Thus the ice melts and flows off the roads and footpaths. Hence in cold countries, the ice frozen on roads and footpaths is melted by spraying salts like sodium chloride or calcium chloride.

Q. 14) Fill in the blank. : Using the relation between the molar masses of solute and colligative properties, it is possible to determine the molecular masses of ………. and ………. solutes from dilute solutions.

  • Using the relation between the molar masses of solute and colligative properties, it is possible to determine the molecular masses of electrolytes and non-electrolyte solutes from dilute solutions.

Q15. Name the law, suggesting the following equations. n1 = n2 … two solutions at same temperature and pressure.



  • The law suggesting the given relation is van’t Hoff-Avogadro’s law

Q16. What is the cause of abnormal colligative properties?

  • Cause: Since colligative properties depend upon the number of particles of the solute, in some cases where the solute associates or dissociates in solution, abnormal results for molecular masses are obtained.

Q17. Name the law suggesting equation p = CRT.

  • The law suggesting equation p = CRT is van\t Hoff general solution equation.

Q18. “Osmotic pressure is more useful to calculate molecular mass than other colligative properties.” Why?

  • In osmotic pressure method, the pressure is measured around room temperature and the molarity of the solution is used instead of molality.
  • The magnitude of osmotic pressure is very large even for dilute solution compared to other colligative properties, this minimizes the change of error in the result.
  • Macromolecules (polymers, proteins) are not stable at high temperature in osmotic pressure macromolecules molecular mass is determined at room temperature where they are stable.
  • Polymers have less solubility. For polymers, osmotic pressure is the best method.
  • Hence osmotic pressure is more useful to calculate molecular mass than other colligative properties.

Q19. Why is freezing point depression of 0.1 M sodium chloride solution is nearly twice than that of 0.1 M glucose solution?



  • Sodium chloride being ionic compound ionizes as (NaCl Na++ Cl) in aqueous solution. The concentration of solute particles, in this case, becomes approximately 0.2 M.
  • Glucose is a covalent compound and does not ionize. Thus the concentration of solute particles remains 0.1 M.
  • A freezing point depression is a colligative property which depends on the number of solute particles in the solution.
  • The number of solute particles in NaCl is twice the concentration of glucose solution. Consequently, freezing point depression of NaCl solution is also approximately twice that of glucose solution.

Q20. What happens when red blood corpuscles are placed in 0.5 % NaCl solution?

  • When placed in 0.5% NaCl solution (hypotonic solution), the RBC’s will swell due to movement of water inside it due to osmosis. It will lead to cell burst.

Q21. Complete the sentence. During the depression of freezing point in a solution, the equilibrium is present in ……….

  • Complete the sentence. During the depression of freezing point in a solution, the equilibrium is present in ……….

Q22. Is it possible to calculate the molar mass of polymers using boiling point or freezing point method? Why?

  • In order to determine the molecular mass of polymers, it is necessary to dissolve the polymer in an appropriate solvent and begin with a dilute solution.
  • In real polymer solutions are solutions of many components. The various components are the polymer species of different molecular masses.
  • Hence it is not possible to calculate the molar mass of polymers using boiling point or freezing point method. However, this methods can be used for polymers having low molar mass.
  • For polymers, osmotic pressure method is used.

Q23. If a grape is placed in water it swells, why?

  • If a grape is placed in water it swells because of osmosis. As the concentration of liquid sap (solution in water) inside grape is more than the concentration of solvent (pure water) outside the grape.
  • Due to osmosis, the solvent (water) moves from low concentration region (from water) to high concentration region (inside grape) till the osmotic pressure on both the sides of the wall of grapes becomes same. Thus water enters the grape and it swells.

Section A – Questions of 1 Mark VSA (which can be combined with sections B / C / D)

Q1. Define the term, solution.

  • A solution is a homogeneous mixture of two or more than two or more components.

Q2. What are molality and molarity?



  • Molality: Molality (m) is defined as a number of moles of solute expressed in kg dissolved in one kg of solvent.
  • Molarity: Molarity (M) is defined as a number of moles of solute dissolved in one litre (or one cubic decimetre) of the solution.

Q3. What is a solid solution?

  • a solid mixture containing a minor component uniformly distributed within the crystal lattice of the major component.
  • Example: Bronze is an alloy consisting primarily of copper, commonly with about 12% tin and often with the addition of other metals (such as aluminium, manganese, nickel or zinc) and sometimes non-metals or metalloids such as arsenic, phosphorus or silicon.

Q4. State Henry’s law.

  • The law states that at a constant temperature, the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas.

S α  P

∴  S = KP



Q5. Define colligative properties.

  • Colligative properties are those properties of dilute solutions that depend only on the number of solute particles ( atoms or molecules, ions or aggregates of molecules) in the solution and not on the nature of solute particles.

Q6. Write the mathematical expression for the lowering of vapour pressure.

  • The mathematical expression for the lowering of vapour pressure is

Δp = p1ox2

Where Δp = Lowering of vapour pressure

p1o = Vapour pressure of the pure solvent

 x2  = Mole fraction of the solute



Q7. Write the mathematical expression for the relative lowering of vapour pressure.

  • The mathematical expression for the relative lowering of vapour pressure is

Where Δp/p1o = Relative lowering of vapour pressure

p1o = Vapour pressure of the pure solvent



 x2  = Mole fraction of the solute

Q8. Define the term lowering of vapour pressure, on the basis of Raoult’s law for a solution of non-volatile solute.

  • On the basis of Raoult’s law, the relative lowering of vapour pres­sure of a solution containing a non­volatile solute is defined as the mole fraction of the solute in the solution.

Q9. Define boiling point.

  • The boiling point of a liquid is defined as the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure.

Q10. Define freezing point.

  • The freezing point of a liquid is defined as a temperature at which the vapour pressure of the solid is equal to the vapour pressure of the liquid.

Q11. How is lowering of vapour pressure is mathematically related to the freezing point?

  • The depression of freezing point is proportional to the lowering of vapour pressure. Thus, mathematically

ΔTf = ∝ p1– P

Where ΔTf = Depression in freezing point

p1o = Vapour pressure of the pure solvent

 P  = Vapour pressure of the solution

Q12. Define osmosis.



  • The spontaneous and unidirectional flow of solvent molecules through a semipermeable membrane, into a solution OR flow of solvent from a solution of low concentration to a solution of higher concentration through a semipermeable membrane, is called osmosis.

Q13. State van’t Hoff- Boyle’s law.

  • At constant temperature the osmotic pressure (π) of a dilute solution is directly proportional to its molar concentration (C) or inversely proportional to volume (V) of the solution. Mathematically

π ∝ C or π ∝ 1/V

Q14. State van’t Hoff-Charles’ law.

  • The concentration remaining constant, the osmotic pressure (π) of a dilute solution is directly proportional to absolute temperature (T) of the solution. Mathematically

π ∝ T

Q15. Write the mathematical expression for van’t Hoff general solution equation.

  • The mathematical expression for van’t Hoff general solution equation is

π = CRT

Where π = osmotic pressure of the solution

C = Concentration of solution

R = Universal gas constant

T = Absolute temperature of the solution



Q16. How is the molar mass of non – volatile solute, related to the colligative property?

  • The colligative properties are inversely proportional to the molecular mass of non-volatile solute

Q17. If a non-volatile solute produces four ions after dissociation, then what will be the change in its molecular mass and colligative property of its solution?

  • The colligative properties of solution like osmotic pressure, elevation of boiling point, and depression of freezing point are much higher than those calculated on the basis of an undissociated single molecule. and the molecular mass is lower.

Q18. State van’t Hoff-Avogadro’s law.

  • Two solutions of equal concentrations of different solutes exert same osmotic pressure at the same temperature OR equal volumes of isotonic solutions contain an equal number of solute particles at the given temperature.

Q19. Write the mathematical expression, showing the relationship between van’t Hoff factor and degree of dissociation.

  • Consider an electrolyte AxBy be dissolved in the solvent. Its dissociation in the solvent is given by

AxBy   \rightleftharpoons  x An+ + Bn-

Then n’ = x + y

The relation between van’t Hoff factor (i) and degree of dissociation (α) is given by

i = 1 + (n’ – 1)α

Q. 20) If an electrolyte HCI is dissolved in water, its value for ΔTf/m= 1.86 K kg mol-1 at 25°C; then what is its value for Kf?

ΔTf   = Kf. m



∴  K= ΔTf/m  = 1.86 K kg mol-1

Q21. What is theoretical colligative property?

  • Theoretical colligative properties are those colligative properties which are inversely proportional to the molecular mass of the solute and there is no variation in them due to association or dissociation.

Q22, Define mole fraction.

  • The mole fraction of any component of a solution is defined as the ratio of the number of moles of that component present in the solution to the total number of moles of all components of the solution.
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2 Comments

  1. nakul bendale

    loved that you created solution for question bank for free of cost you could have sold your solution and would have earned money .but thanks for help;

  2. Thanks a lot for the solutions. Really helped a lot. God bless you 🙂

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