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Chapter 12: Problem 23

A solution prepared by dissolving \(0.8220 \mathrm{~g}\) of glucose, a nonelectrolyte, in enough water to produce \(200.0 \mathrm{~mL}\) of solution has an osmotic pressure of \(423.1 \mathrm{~mm} \mathrm{Hg}\) at \(298 \mathrm{~K}\). What is the molecular weight of glucose?

Short Answer

Expert verified
The molecular weight of glucose is approximately 180 g/mol.

Step by step solution

01

Understand the Osmotic Pressure Formula

The formula for osmotic pressure (\(\pi\)) is given by \[\pi = iCRT\], where \(i\) is the van 't Hoff factor, \(C\) is the molarity of the solution, \(R\) is the ideal gas constant (\(0.0821 \text{ L atm } \text{mol}^{-1} \text{K}^{-1}\)), and \(T\) is the temperature in Kelvin. For a nonelectrolyte like glucose, \(i=1\).
02

Convert Units

Convert osmotic pressure from mm Hg to atm using the conversion \(1 \text{ atm} = 760 \text{ mm Hg}\). So, \(423.1 \text{ mm Hg} = \frac{423.1}{760} \text{ atm}\). Similarly, convert volume to liters: \(200.0 \text{ mL} = 0.2000 \text{ L}\).
03

Substitute into Osmotic Pressure Equation

Rearrange the osmotic pressure equation to solve for molarity (\(C\)): \[C = \frac{\pi}{RT}\]. Substitute the values \(\pi = \frac{423.1}{760} \text{ atm}\), \(R = 0.0821 \text{ L atm } \text{mol}^{-1} \text{K}^{-1}\), and \(T = 298 \text{ K}\) into the equation.
04

Calculate Moles of Solute

From the molarity \(C\), which has units of moles per liter, multiply by the volume in liters (\(0.2000 \text{ L}\)) to get the moles of solute present in the solution.
05

Determine the Molecular Weight

Calculate the molar mass (molecular weight) of glucose by dividing the mass of glucose in the solution (\(0.8220 \text{ g}\)) by the moles of glucose determined in Step 4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Weight Calculation
Calculating the molecular weight of a substance like glucose involves determining how many grams are present per mole of the substance. To achieve this, one must first determine the number of moles of glucose in the solution. Moles can be calculated using the formula for molarity:
  • Molarity (C) = Moles of solute / Volume of solution in liters
Once the moles are found, simply divide the given mass of glucose (\(0.8220 \, \text{g}\) in this case) by the number of moles to find the molecular weight. This approach allows converting measurable quantities into molecular form, giving insight into the molecular makeup of the solute.
Glucose as Nonelectrolyte
Glucose is a nonelectrolyte, which means it doesn't dissociate into ions when dissolved in water. Unlike electrolytes, nonelectrolytes remain intact and don't enhance the electrical conductivity of the solution. This characteristic simplifies calculations since the van 't Hoff factor (\(i\)) for glucose is 1. This factor indicates that glucose behaves as one whole molecule, making it crucial for osmotic pressure calculations, where dissociation does not affect the pressure measurement. This streamlined behavior helps educators simplify the explanation of osmotic properties in nonelectrolytic solutions.
Van't Hoff Factor
The van 't Hoff factor (\(i\)) reflects the degree of dissociation or ionization of a solute in a solution. For glucose and other nonelectrolytes, \(i = 1\). This means no dissociation occurs, simplifying calculations involving the osmotic pressure equation:
  • \[\pi = iCRT\]
The understanding of this factor is critical for different compounds, as electrolytes, like salt, would have an\(i\) greater than 1. For educators, illustrating how the van 't Hoff factor changes with different solutes can reinforce understanding of solution behavior and its impact on properties like boiling point elevation and osmotic pressure.
Molarity in Solution Chemistry
Molarity is a fundamental concept in solution chemistry, defined as the number of moles of solute per liter of solution. It provides a quantifiable measure of concentration. In this exercise, knowing the molarity is essential for applying the formula for osmotic pressure, rearranging it to:
  • \[C = \frac{\pi}{RT}\]
Calculating molarity involves converting volume to liters and solving for moles through the osmotic pressure equation. Understanding molarity's role in chemical equations helps students grasp how solute concentration affects solution properties, allowing them to apply similar principles to various chemical reactions and experiments.

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