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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 (C12H22O11C_{12}H_{22}O_{11}) per litre. The molecular mass of protein is a×10xa \times 10^xu. Find the value of a + x.

Answer

8

Explanation

Solution

  1. Isotonic Solutions: Isotonic solutions have equal molar concentrations.
  2. Molar Concentration Formula: Molar concentration (CC) is given by C=wMVC = \frac{w}{M \cdot V}, where ww is the mass, MM is the molar mass, and VV is the volume.
  3. Equality for Isotonic Solutions: For isotonic solutions with the same volume VV, their molar concentrations are equal: w1M1V=w2M2V\frac{w_1}{M_1 \cdot V} = \frac{w_2}{M_2 \cdot V}. This simplifies to w1M1=w2M2\frac{w_1}{M_1} = \frac{w_2}{M_2}.
  4. Molar Mass of Sucrose: The molar mass of sucrose (C12H22O11C_{12}H_{22}O_{11}) is calculated as (12×12.011)+(22×1.008)+(11×15.999)342(12 \times 12.011) + (22 \times 1.008) + (11 \times 15.999) \approx 342 g/mol.
  5. Given Values:
    • Mass of protein (w1w_1) = 200 g/L
    • Mass of sucrose (w2w_2) = 1.71 g/L
    • Molar mass of sucrose (M2M_2) = 342 g/mol
  6. Substituting Values: Substitute these values into the simplified equation: 200 gMprotein=1.71 g342 g/mol\frac{200 \text{ g}}{M_{protein}} = \frac{1.71 \text{ g}}{342 \text{ g/mol}}.
  7. Solving for Protein's Molecular Mass: Mprotein=200 g×342 g/mol1.71 gM_{protein} = \frac{200 \text{ g} \times 342 \text{ g/mol}}{1.71 \text{ g}} Mprotein=200×3421.71 g/molM_{protein} = \frac{200 \times 342}{1.71} \text{ g/mol} To simplify, multiply the numerator and denominator by 100: Mprotein=200×342×100171 g/molM_{protein} = \frac{200 \times 342 \times 100}{171} \text{ g/mol} Recognize that 342=2×171342 = 2 \times 171: Mprotein=200×(2×171)×100171 g/molM_{protein} = \frac{200 \times (2 \times 171) \times 100}{171} \text{ g/mol} Cancel out 171: Mprotein=200×2×100 g/mol=40000 g/molM_{protein} = 200 \times 2 \times 100 \text{ g/mol} = 40000 \text{ g/mol}.
  8. Scientific Notation: The molecular mass of the protein is given as a×10xa \times 10^x u. So, 40000 u=a×10x40000 \text{ u} = a \times 10^x u.
  9. Expressing in Scientific Notation: Express 40000 in scientific notation, where 1a<101 \le a < 10: 40000=4×10440000 = 4 \times 10^4.
  10. Identifying a and x: Therefore, a=4a = 4 and x=4x = 4.
  11. Final Calculation: The value of a+xa + x is 4+4=84 + 4 = 8.