Math Problem Statement

A solution has 12% of sucrose by weight (molar mass 342.3 g mol-1). The osmotic pressure of the red blood cell is 5.94 bar at 25 °C, how much (in terms of volume) of the 12% sucrose solution do you need to make 1 L of isotonic (isosmotic) solution for the red blood cell?

Solution

To find the volume of the 12% sucrose solution needed to make 1 L of an isotonic solution for the red blood cell, we can apply the concept of osmotic pressure and the van 't Hoff equation for osmotic pressure.

The van 't Hoff equation for osmotic pressure (Π\Pi) is: Π=iCRT\Pi = i \cdot C \cdot R \cdot T where:

  • Π\Pi is the osmotic pressure in bar,
  • ii is the van 't Hoff factor (1 for sucrose, a non-electrolyte),
  • CC is the molarity of the solution (mol/L),
  • RR is the gas constant (0.0831L bar K1mol10.0831 \, \text{L bar K}^{-1} \text{mol}^{-1}),
  • TT is the temperature in Kelvin.

Given values:

  • Osmotic pressure (Π\Pi) = 5.94 bar
  • Temperature T=25C=298KT = 25^\circ \text{C} = 298 \, \text{K}

Step 1: Calculate the required molarity (C)

Rearrange the van 't Hoff equation to solve for CC: C=ΠiRTC = \frac{\Pi}{i \cdot R \cdot T} Substitute values: C=5.941×0.0831×298C = \frac{5.94}{1 \times 0.0831 \times 298} C0.241mol/LC \approx 0.241 \, \text{mol/L}

Step 2: Determine the amount of sucrose in 1 L of solution

Since we need a 1 L isotonic solution with a molarity of 0.241mol/L0.241 \, \text{mol/L}, the moles of sucrose required are: moles of sucrose=0.241mol\text{moles of sucrose} = 0.241 \, \text{mol}

Step 3: Convert moles to mass of sucrose

Using the molar mass of sucrose (342.3g/mol342.3 \, \text{g/mol}): mass of sucrose=0.241×342.3\text{mass of sucrose} = 0.241 \times 342.3 mass of sucrose82.5g\text{mass of sucrose} \approx 82.5 \, \text{g}

Step 4: Calculate the volume of the 12% sucrose solution needed

The 12% sucrose solution means there are 12 g of sucrose in every 100 g of solution. To find the total mass of solution that contains 82.5 g of sucrose: 82.5g sucrose12g sucrose per 100 g solution=82.50.12\frac{82.5 \, \text{g sucrose}}{12 \, \text{g sucrose per 100 g solution}} = \frac{82.5}{0.12} 687.5g of solution\approx 687.5 \, \text{g of solution}

Assuming the density of the solution is approximately 1g/mL1 \, \text{g/mL} (a reasonable assumption for dilute solutions), the volume of the solution needed is roughly: 687.5mL687.5 \, \text{mL}

Answer

To make a 1 L isotonic solution for the red blood cell, you need approximately 687.5 mL of the 12% sucrose solution.

Would you like further clarification on any steps?

Here are some related questions to deepen understanding:

  1. How would this calculation change if the sucrose solution concentration were different?
  2. What assumptions are made about the solution's properties in this calculation?
  3. How would temperature variations affect the volume required?
  4. How does the osmotic pressure of different solutes compare for the same molarity?
  5. What is the importance of isotonic solutions in biological applications?

Tip: When working with osmotic pressure, double-check the units of RR and Π\Pi for consistency.

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Math Problem Analysis

Mathematical Concepts

Osmotic Pressure
Molarity
Dilution
Unit Conversions

Formulas

\Pi = i \cdot C \cdot R \cdot T
C = \frac{\Pi}{i \cdot R \cdot T}
Mass = Moles \times Molar Mass
Volume = Mass / Density

Theorems

Van 't Hoff Law for Osmotic Pressure

Suitable Grade Level

Grades 11-12