Osmotic Pressure
Hardchemistry
A solution containing 0.5 g of a protein in 100 mL of solution has an osmotic pressure of 1.23 × 10⁻³ atm at 300 K. The molar mass of the protein is: (R = 0.082 L atm K⁻¹ mol⁻¹)
Select the correct option:
Solution
Incorrect! Answer:
100000 g/mol
Osmotic Pressure (π) is defined by the formula: π=VnRT=MVwRT
- Given:
- π=1.23×10−3 atm
- w=0.5 g, V=100 mL=0.1 L
- R=0.082 L atm K−1 mol−1, T=300 K
- Calculation: M=πVwRT=1.23×10−3×0.10.5×0.082×300 M=1.23×10−412.3=105=100,000 g/mol.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
More osmotic pressure Practice Questions
About This Question
- Subject
- chemistry
- Chapter
- solutions
- Topic
- osmotic pressure
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
100000 g/mol
Osmotic Pressure (π) is defined by the formula: π=VnRT=MVwRT
- Given:
- π=1.23×10−3 atm
- w=0.5 g, V=100 mL=0.1 L
- R=0.082 L atm K−1 mol−1, T=300 K
- Calculation: M=πVwRT=1.23×10−3×0.10.5×0.082×300 M=1.23×10−412.3=105=100,000 g/mol.
This hard difficulty chemistry question is from the chapter solutions, covering the topic of osmotic pressure. It appeared in the 2025 exam.
Looking for more practice? Explore all chemistry questions or browse solutions questions on RankGuru.