Question
Download Solution PDFThe molality (in mol kg-1) of 1 mole of solute in 50 g of solvent is
This question was previously asked in
AIIMS BSc NURSING 2024 Memory-Based Paper
Answer (Detailed Solution Below)
Option 2 : 20
Free Tests
View all Free tests >
AIIMS BSc NURSING 2024 Memory-Based Paper
6.4 K Users
100 Questions
100 Marks
120 Mins
Detailed Solution
Download Solution PDFCONCEPT:
Molality (m)
- Molality (m) is a measure of the concentration of a solute in a solution.
- It is defined as the number of moles of solute per kilogram of solvent.
- The formula for molality is:
m = (moles of solute) / (mass of solvent in kg)
EXPLANATION:
- Given data:
- Moles of solute = 1 mole
- Mass of solvent = 50 g
- Convert mass of solvent to kilograms:
- 50 g = 50 / 1000 kg = 0.05 kg
- Using the molality formula:
- m = (moles of solute) / (mass of solvent in kg)
- = 1 mole / 0.05 kg
- = 20 mol kg-1
Therefore, the molality of 1 mole of solute in 50 g of solvent is 20 mol kg-1.
Other Options Explanation:
- Option 1 (10 mol kg-1):
- This would be the result if the mass of the solvent was 100 g (0.1 kg) instead of 50 g.
- m = 1 mole / 0.1 kg = 10 mol kg-1
- Option 3 (30 mol kg-1):
- This would be the result if the mass of the solvent was approximately 33.33 g (0.03333 kg).
- m = 1 mole / 0.03333 kg ≈ 30 mol kg-1
- Option 4 (40 mol kg-1):
- This would be the result if the mass of the solvent was 25 g (0.025 kg).
- m = 1 mole / 0.025 kg = 40 mol kg-1
Last updated on Jun 17, 2025
-> The AIIMS BSc Nursing (Hons) Result 2025 has been announced.
->The AIIMS BSc Nursing (Hons) Exam was held on 1st June 2025
-> The exam for BSc Nursing (Post Basic) will be held on 21st June 2025.
-> AIIMS BSc Nursing Notification 2025 was released for admission to the 2025 academic session.