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Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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**Title: Osmosis Experiment: Glucose Solution in Dialysis Tubing**

**Experiment Objective:**
For an experiment on osmosis, some dialysis tubing (a semi-permeable membrane) is filled with 750 mL of a glucose (C6H12O6, non-electrolyte) solution (glucose cannot pass through the membrane). If the reaction is carried out at 36.5°C and creates an osmotic pressure of 1.75 atm, what would be the concentration of the glucose solution?

**Key Calculations and Formulas:**
1. **Formula for Osmotic Pressure (π):**
   \[\pi = MRT\]
   Where,
   - \(\pi\) = osmotic pressure (in atm)
   - \(M\) = molarity of the solution (in mol/L)
   - \(R\) = gas constant (0.0821 L·atm/(K·mol))
   - \(T\) = temperature (in Kelvin)

2. **Converting Celsius to Kelvin:**
   \[
   T(K) = T(°C) + 273
   \]
   For 36.5°C: 
   \[
   36.5 + 273 = 309.5 \, K
   \]

3. **Rearranging the Osmotic Pressure Formula to Solve for Molarity (M):**
   \[
   M = \frac{\pi}{RT}
   \]

**Given Data:**
- Osmotic Pressure (π) = 1.75 atm
- Gas Constant (R) = 0.0821 L·atm/(K·mol)
- Temperature (T) = 36.5°C = 309.5 K 

**Calculation Steps:**
1. Substitute the values into the rearranged formula:
   \[
   M = \frac{1.75 \, \text{atm}}{0.0821 \, \text{L·atm/(K·mol)} \times 309.5 \, \text{K}}
   \]

2. Perform the calculation:
   \[
   M = \frac{1.75}{0.0821 \times 309.5}
   \]

3. Simplify the calculation to find the molarity (M):
   \[
   M = 0.0686 \, \text{mol/L}
   \]
Transcribed Image Text:**Title: Osmosis Experiment: Glucose Solution in Dialysis Tubing** **Experiment Objective:** For an experiment on osmosis, some dialysis tubing (a semi-permeable membrane) is filled with 750 mL of a glucose (C6H12O6, non-electrolyte) solution (glucose cannot pass through the membrane). If the reaction is carried out at 36.5°C and creates an osmotic pressure of 1.75 atm, what would be the concentration of the glucose solution? **Key Calculations and Formulas:** 1. **Formula for Osmotic Pressure (π):** \[\pi = MRT\] Where, - \(\pi\) = osmotic pressure (in atm) - \(M\) = molarity of the solution (in mol/L) - \(R\) = gas constant (0.0821 L·atm/(K·mol)) - \(T\) = temperature (in Kelvin) 2. **Converting Celsius to Kelvin:** \[ T(K) = T(°C) + 273 \] For 36.5°C: \[ 36.5 + 273 = 309.5 \, K \] 3. **Rearranging the Osmotic Pressure Formula to Solve for Molarity (M):** \[ M = \frac{\pi}{RT} \] **Given Data:** - Osmotic Pressure (π) = 1.75 atm - Gas Constant (R) = 0.0821 L·atm/(K·mol) - Temperature (T) = 36.5°C = 309.5 K **Calculation Steps:** 1. Substitute the values into the rearranged formula: \[ M = \frac{1.75 \, \text{atm}}{0.0821 \, \text{L·atm/(K·mol)} \times 309.5 \, \text{K}} \] 2. Perform the calculation: \[ M = \frac{1.75}{0.0821 \times 309.5} \] 3. Simplify the calculation to find the molarity (M): \[ M = 0.0686 \, \text{mol/L} \]
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