9th Chapter Solved Exercise MCQs of FSC First Year Chemistry
The 9th chapter Solved Exercise MCQs of FSC 1st-year Chemistry offers a comprehensive collection of multiple-choice questions with answers. Each question is accompanied by a short explanation to clarify the correct choice. These solutions aim to assist students in exam preparation by simplifying complex concepts and building a strong foundation for success.
9th chapter solved MCQs with explanation
1. Molarity of pure water is:
(a) 1
(b) 18
(c) 55.5
(d) 6
Explanation: The correct answer is (c). Because molarity means the number of moles of a substance in 1 liter of solution. For water, 1 mole weighs 18 grams.
- In 1 liter of pure water, there are 1000 grams.
- To find out how many moles are in 1000 grams:
Number of moles = 1000 grams / 18 grams per mole = 55.5 moles
2. 18 g glucose is dissolved in 90 g of water. The relative lowering of vapor pressure is equal to:
(a) 1/5
(b) 5.1
(c) 1/51
(d) 6
Explanation: The correct answer is (c). Because:
- Formula to use: ΔP / P0 = Number of moles of solute / Number of moles of solvent
- Find the number of moles:
-Glucose (solute):
Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol.
Mass of glucose = 18 g.
Number of moles of glucose = 18/180 = 0.1 mol.
-Water (solvent):
Molar mass of water (H₂O) = 18 g/mol.
Mass of water = 90 g.
Number of moles of water = 90/18 = 5 mol.
- Calculate the relative lowering of vapor pressure: ΔP/P0 = 0.1 / 5 = 0.02
- Convert to fraction: 0.02 can be written as 1/50. Hence the answer is closest to 1/51.
3. A solution of glucose is 10% w/v. The volume in which 1 g mole of it is dissolved will be:
(a) 1 dm³
(b) 1.8 dm³
(c) 200 cm³
(d) 900 cm³
Explanation: The correct answer is (b). Because for a 10% w/v glucose solution, 10 grams of glucose is in 100 mL of solution.
- To find out how much volume is needed for 1 mole (180 grams) of glucose:
Volume = 180 g/10 g x 100 mL = 1800 mL = 1.8 dm3
4. An aqueous solution of ethanol in water may have vapor pressure:
(a) equal to that of water
(b) equal to that of ethanol
(c) more than that of water
(d) less than that of water
Explanation: The correct answer is (d). Because when you mix ethanol with water, the vapor pressure of the mixture is less than that of pure water. This is because ethanol and water interact with each other, making it harder for the ethanol molecules to escape into the air compared to when ethanol is alone. Thus, even though ethanol has a higher vapor pressure than water on its own, mixing it with water lowers the overall vapor pressure of the solution.
5. An azeotropic mixture of two liquids boils at a lower temperature than either of them when:
(a) it is saturated
(b) it shows positive deviation from Raoult’s law
(c) it shows negative deviation from Raoult’s law
(d) it is metastable
Explanation: The correct answer is (b). Because intermolecular forces between solute and solvent are weaker than interactions between solute and solvent when a solution deviates from Raoult’s law. The total vapor pressure of the solution is greater than the total vapor pressure of an ideal solution. Consequently, each component’s vapor pressure is greater than expected by Raoult’s law. Therefore, the azeotropic mixture of two liquids boils at a lower temperature than either liquid alone.
6. In an azeotropic mixture showing positive deviation from Raoult’s law, the volume of the mixture is:
(a) slightly more than the total volume of the components
(b) slightly less than the total volume of the components
(c) equal to the total volume of the components
(d) none of these
Explanation: The correct answer is (a). Because the weaker intermolecular forces in the mixture lead to a larger volume compared to the sum of the individual component volumes.
7. Which of the following solutions has the highest boiling point?
(a) 5.85% solution of sodium chloride
(b) 18.0% solution of glucose
(c) 6.0% solution of urea
(d) All have the same boiling point
Explanation: The correct answer is (a). Because:
- Sodium chloride (NaCl) dissociates into 2 ions (Na⁺ and Cl⁻) in solution. Thus, a 5.85% NaCl solution will have a higher boiling point due to the presence of more particles compared to the other solutions.
- Glucose and urea do not dissociate into ions; they remain as single molecules in solution. Therefore, their contributions to boiling point elevation are lower compared to NaCl.
8. Two solutions of NaCl and KCl are prepared separately by dissolving the same amount of the solute in water. Which of the following statements is true for these solutions?
(a) KCl solution will have a higher boiling point than NaCl solution
(b) Both the solutions have different boiling points
(c) KCl and NaCl solutions possess the same vapor pressure
(d) KCl solution possesses a lower freezing point than NaCl solution
Explanation: The correct answer is (b). Because even though both salts dissolve into two ions, they have different effects on the boiling point of the solution. NaCl and KCl interact differently with water, so their solutions won’t boil at the same temperature.
9. The molal boiling point constant is the ratio of the elevation in boiling point to:
(a) molarity
(b) mole fraction of solvent
(c) molality
(d) mole fraction of solute
Explanation: The correct answer is (c). Because the molal boiling point constant (Kb) is used to calculate the elevation in boiling point of a solvent when a solute is added. The formula for the boiling point elevation (ΔTb) is:
ΔTb = Kb x m
Where m is the molality of the solution (moles of solute per kilogram of solvent). Therefore, the molal boiling point constant (Kb) is the ratio of the elevation in boiling point to the molality of the solution.
10. Colligative properties are the properties of:
(a) dilute solutions which behave as nearly ideal solutions
(b) concentrated solutions which behave as nearly non-ideal solutions
(c) both (i) and (ii)
(d) neither (i) nor (ii)
Explanation: The correct answer is (a). Because colligative properties depend on the number of solute particles, not their nature. These properties are most accurate in dilute solutions because the solute particles are spread out enough to minimize interactions between them, allowing the solution to behave more ideally. In concentrated solutions, deviations from ideal behavior occur due to significant solute-solute interactions
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