Problem 82 - Chemistry: The Central Science in SI Units, Global Edition

Question

A dilute aqueous solution of fructose in water is formed by dissolving \(1.25 \mathrm{~g}\) of the compound in water to form \(0.150 \mathrm{~L}\) of solution. The resulting solution has an osmotic pressure of \(112.8 \mathrm{kPa}\) at \(20^{\circ} \mathrm{C}\). Assuming that the organic compound is a nonelectrolyte, what is its molar mass?

Step-by-Step Solution

Answer

The molar mass of fructose is approximately \(186.9 \, \text{g/mol}\), calculated using the given mass, volume, osmotic pressure, and temperature with the osmotic pressure equation.

Step 1: Convert given values to appropriate units

We need to make sure that all given values have the appropriate units. Mass of fructose: 1.25 g Volume of solution: 0.150 L Osmotic pressure: 112.8 kPa Temperature: 20°C 1. Convert kPa to atm: 1 atm = 101.325 kPa Osmotic pressure in atm: \(112.8 \,\text{kPa} × \frac{1 \,\text{atm}}{101.325 \,\text{kPa}} \approx 1.113 \,\text{atm}\) 2. Convert Celsius to Kelvin: \(T = 20°C + 273.15 = 293.15 \,\text{K}\) Values in appropriate units: Osmotic pressure: 1.113 atm Temperature: 293.15 K

Step 2: Use the osmotic pressure equation to find molar concentration

We will use the osmotic pressure equation to solve for the molar concentration (c): \(Π = cRT\) where: Π = 1.113 atm R = 0.0821 L atm / K mol T = 293.15 K Rearrange the equation to solve for c: \(c = \frac{Π}{RT}\) Now substitute the values: \(c = \frac{1.113 \,\text{atm}}{(0.0821 \,\text{(L atm)}/\text{(K mol})) \times 293.15 \,\text{K}} \approx 0.0446 \frac{\text{mol}}{\text{L}}\) Molar concentration: 0.0446 mol/L

Step 3: Find the molar mass of fructose using the molar concentration

We can calculate the moles of fructose using the molar concentration and the volume of the solution: Moles of fructose = Concentration × Volume Moles of fructose = (0.0446 mol/L) × (0.150 L) ≈ 0.00669 mol Now we will use the mass and moles of fructose to find its molar mass: Molar mass = Mass / Moles Molar mass = (1.25 g) / (0.00669 mol) ≈ 186.9 g/mol Molar mass of fructose: 186.9 g/mol

Key Concepts

Molar Mass Calculation

Molar mass is a crucial concept in chemistry, especially when working with solutions. It helps you understand the weight of a specific amount of substance. In essence, molar mass is the mass of a given substance divided by the amount of substance in moles. It’s expressed in grams per mole (g/mol).

The calculation involves determining how many moles of a compound are in a given mass. For instance, if you have a solution with a known mass of solute, such as 1.25 grams of fructose, and you find that there are approximately 0.00669 moles of the solute, you use the formula:

• Molar Mass = \( \frac{\text{Mass (g)}}{\text{Moles (mol)}} \)
• Molar Mass = \( \frac{1.25 \, \text{g}}{0.00669 \, \text{mol}} \approx 186.9 \, \text{g/mol} \)

This calculation is essential for converting between mass and moles, which is vital for understanding various properties of the solution.

Nonelectrolyte Solution

Understanding how nonelectrolyte solutions behave is important in chemistry. A nonelectrolyte is a substance that, when dissolved in water, does not dissociate into ions. Unlike electrolytes, nonelectrolytes do not conduct electricity in a solution.

When working with nonelectrolyte solutions, the primary concern is usually about their physical properties, like osmotic pressure, which are influenced by the concentration of molecules rather than ions. In our case, fructose is considered a nonelectrolyte in the solution, meaning that it retains its molecular structure in water. Since it doesn't produce ions, the calculation for osmotic properties does not need to account for ion effects.

• Solute doesn't dissociate in the solution.
• Osmotic pressure depends on the number of molecules, not ions.

This characteristic simplifies calculations and is crucial when studying properties like osmotic pressure in solutions.

Molar Concentration

Molar concentration, often referred to as molarity, is a way of expressing the concentration of a solution. It tells you how many moles of solute are present per liter of solution, commonly indicated with the symbol \( c \).

In our context, using the osmotic pressure equation \( \Pi = cRT \), which links osmotic pressure (\( \Pi \)), molar concentration (\( c \)), the gas constant (\( R \)), and temperature (\( T \)), we solve for \( c \):

• Rearrange the equation: \( c = \frac{\Pi}{RT} \)
• Substitute the known values to find \( c = \frac{1.113 \, \text{atm}}{(0.0821 \, \text{L atm}/\text{K mol}) \times 293.15 \, \text{K}} \approx 0.0446 \, \text{mol/L} \)

Molar concentration provides an understanding of solute quantity in a solution and is a foundational concept when working with solution chemistry. This value informs us about the number of molecules present, which directly impacts properties like osmotic pressure.