Tiêu đề 2. Solutions.pdf ✅

Solutions Class 12 Important Questions Very Short Answer Type Question 1. Differentiate between molarity and molality of a solution. (All India 2010) Answer: The distinction between molarity and molality. Molarity: It is the number of moles of solute dissolved in 1 litre of solution. It is temperature dependent. M = ω×1000 mol.mass ×V Molality : It is the number of moles of solute dissolved in 1 kg of the solvent. m = ω×1000M2×W The relationship between molarity and molality is m = Md−MM21000 When molality = molarity, we get, 1 = 1d−MM21000 or d – MM ̇21000 = 1 ∴ d = 1+MM21000 Molarity is temperature dependent while molarity is not. For very dilute solution, the factor MM2/1000 can be neglected in comparison to 1. Hence molality will be same to molarity when density d = 1. Molality is independent of temperature, whereas molarity is a function of temperature because volume depends on temperature and mass does not. Question 2. What type of semiconductor is obtained when silicon is doped with arsenic? (Delhi 2010) Answer: n-type semiconductor. Question 3. What is meant by ‘reverse osmosis’? (All India 2011) Answer: If a pressure higher than the osmotic pressure is applied on the solution, the solvent will flow from the solution into the pure solvent through semipermeable membrane. This process is called reverse osmosis (R.O.). Question 4. What are isotonic solutions? (Delhi 2014) Answer: An isotonic solution is a kind of solution with the same salt concentration as blood and cells. Those solutions which are exerting same osmotic pressure under similar conditions (For example 0.9% NaCl solution by mass volume is Isotonic with human blood). Question 5. Some liquids on mixing form ‘azeotropes. What are ‘azeotropes’? (Delhi 2014)
Answer: The liquid mixture having a definite composition and boiling like a pure liquid without change in composition is called as azeotrope. Question 6. What type of intermolecular attractive interaction exists in the pair of methanol and acetone? (Delhi 2014) Answer: Solute-solvent dipolar interactions exist in the pair of methanol and acetone. Question 7. Out of BaCl2 and KCl, which one is more effective in causing coagulation of a negatively charged colloidal Sol? Give reason. (Delhi 2015) Answer: BaCl2 is more effective in causing coagulation because it has double +ve charge than K+. Solutions Class 12 Important Questions Short Answer Type – I [SA- I] Question 8. Differentiate between molality and molarity of a solution. What is the effect of change in temperature of a solution on its molality and molarity? (Delhi 2009) Answer: Distinction between molarity and molality. Molarity: It is the number of moles of solute dissolved in 1 litre of solution. It is temperature dependent. M = ω×1000 mol. mass ×V Molality: It is the number of moles of solute dissolved in 1 kg of the solvent. m = ω×1000M2×W The relationship between molarity and molality is m = Md−MM21000 When molality = molarity, we get, 1 = 1d−MM21000 or d – MM ̇21000 = 1 ∴ d = 1+MM21000 Molarity is temperature dependent while molarity is not. For very dilute solution, the factor MM2/1000 can be neglected in comparison to 1. Hence molality will be same to molarity when density d = 1. Molality is independent of temperature, whereas molarity is a function of temperature because volume depends on temperature and mass does not. Question 9. Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example for each type. (Delhi 2010)
Answer: Non-ideal solutions exhibit Negative deviation from Raoult’s law: For any composition of the non-ideal solution, the partial vapour pressure of each component and total vapour pressure of the solution is less than expected from Raoult’s law. Such solutions show negative deviation. Example: Mixture of CHCl3 and acetone. Non-ideal solutions show positive deviations from Raoult’s law on mixing of two volatile components of the solution. Example: Mixture of acetone and benzene solutions show positive deviation, Question 10. Define the terms, ‘osmosis’ and ‘osmotic pressure’. What is the advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses of solutes in solutions? (All India 2010) Answer: Osmosis : The net spontaneous flow of the solvent molecules from the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semipermeable membrane is called osmosis. Osmotic pressure : The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane is called the osmotic pressure. The osmotic pressure method has the advantage that it uses molarities instead of molalities and it can be measured at room temperature. Question 11. A 1.00 molal aqueous solution of trichloroacetic acid (CCl3COOH) is heated to its boiling point. The solution has the boiling point of 100.18°C. Determine the van’t Hoff factor for trichloroacetic acid. (Kb for water = 0.512 K kg mol-1) (Delhi 2012) Answer: As ΔTb= iKbm (100.18 – 100) °C = i × 0.512 K kg mol-1 × 1 m 0.18 K = i × 0.512 K kg mol-1 × 1 m ∴ i = 0.3 Question 12. Define the following terms : (i) Mole fraction
(ii) Isotonic solutions (iii) van’t Hoff factor (iv) Ideal solution (Delhi 2012) Answer: (i) Mole fraction : Mole fraction is the ratio of number of moles of one component to the total number of moles in a mixture. (ii) Isotonic solution : Two solutions having same osmotic pressure at a given temperature are called Isotonic solutions. (iii) van’t Hoff factor : van’t Hoff factor is expressed as : i = normal molar mass abnormal molar mass (iv) Ideal solution : The solution which obeys Raoult’s law under all conditions is known as an ideal solution. Question 13. Explain why aquatic species are more comfortable in cold water rather than in warm water. (Comptt. Delhi 2012) Answer: Aquatic species need dissolved oxygen for breathing. As solubility of gases decreases with increase of temperature, less oxygen is available in summer in the lake. Hence the aquatic species feel more comfortable in winter (low temperature) when the solubility of oxygen is higher. Question 14. State Raoult’s law. How is it formulated for solutions of non-volatile solutes? (Comptt. Delhi 2012) Answer: Raoult’s Law: Raoult’s Law states that “for a solution of volatile liquids, the partial vapour of each component in the solution is directly proportional to its mole fraction”. Thus for component 1 : p1 = p1 0 X1 where [p1 0 is vapour pressure of pure component 1] For component 2 : p2 = P2 0 X2 According to Dalton’s law of partial pressure PTotal= P2 + P2 ⇒ PT = p1 0X1 + p2 0X2 ⇒ PT = p1 0(1 – X2) + p2 0X2 ⇒ PT = p1 0+ (p2 0 – p1 0)X2 Question 15. State Henry’s law and mention two of its important applications. (Comptt. All India 2012) Answer: Henry’s law: Henry’s law states that “The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution “, Applications of Henry’s law: • To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high pressure.