Question
Question: A solution containing 200 g of a protein per litre is isotonic with 1.71g of sucrose ($C_{12}H_{22}O...
A solution containing 200 g of a protein per litre is isotonic with 1.71g of sucrose (C12H22O11) per litre. The molecular mass of protein is a×10xu. Find the value of a + x.

Answer
8
Explanation
Solution
- Isotonic Solutions: Isotonic solutions have equal molar concentrations.
- Molar Concentration Formula: Molar concentration (C) is given by C=M⋅Vw, where w is the mass, M is the molar mass, and V is the volume.
- Equality for Isotonic Solutions: For isotonic solutions with the same volume V, their molar concentrations are equal: M1⋅Vw1=M2⋅Vw2. This simplifies to M1w1=M2w2.
- Molar Mass of Sucrose: The molar mass of sucrose (C12H22O11) is calculated as (12×12.011)+(22×1.008)+(11×15.999)≈342 g/mol.
- Given Values:
- Mass of protein (w1) = 200 g/L
- Mass of sucrose (w2) = 1.71 g/L
- Molar mass of sucrose (M2) = 342 g/mol
- Substituting Values: Substitute these values into the simplified equation: Mprotein200 g=342 g/mol1.71 g.
- Solving for Protein's Molecular Mass: Mprotein=1.71 g200 g×342 g/mol Mprotein=1.71200×342 g/mol To simplify, multiply the numerator and denominator by 100: Mprotein=171200×342×100 g/mol Recognize that 342=2×171: Mprotein=171200×(2×171)×100 g/mol Cancel out 171: Mprotein=200×2×100 g/mol=40000 g/mol.
- Scientific Notation: The molecular mass of the protein is given as a×10x u. So, 40000 u=a×10x u.
- Expressing in Scientific Notation: Express 40000 in scientific notation, where 1≤a<10: 40000=4×104.
- Identifying a and x: Therefore, a=4 and x=4.
- Final Calculation: The value of a+x is 4+4=8.