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A solution of \(6.85 \mathrm{~g}\) of a carbohydrate in \(100.0 \mathrm{~g}\) of water has a density of \(1.024 \mathrm{~g} / \mathrm{mL}\) and an osmotic pressure of 4.61 atm at \(20.0^{\circ} \mathrm{C}\). Calculate the molar mass of the carbohydrate.

Short Answer

Expert verified
The molar mass of the carbohydrate is approximately 347.21 g/mol.

Step by step solution

01

Define the Osmotic Pressure Formula

The formula for osmotic pressure ( \(\pi\) ) of a solution is given by \(\pi = \frac{n}{V}RT\), where \(n\) is the number of moles of solute, \(V\) is the volume of the solution in liters, \(R\) is the ideal gas constant \(0.0821 \frac{L·atm}{K·mol}\), and \(T\) is the temperature in Kelvin.The osmotic pressure given is \(4.61 \, \text{atm}\). The temperature is \(20.0 \degree C\), but we need it in Kelvin, so add 273.15 to get \(293.15 \, \text{K}\).
02

Calculate the Mass of the Solution

The total mass of the solution is the mass of the solute plus the mass of the solvent: \(6.85 \, \text{g} + 100.0 \, \text{g} = 106.85 \, \text{g}\).
03

Find the Volume of the Solution

Use the density formula to find the volume: \(\text{Density} = \frac{\text{mass}}{\text{volume}}\). Rearrange to solve for volume: \(\text{volume} = \frac{\text{mass}}{\text{Density}}\).Substitute the values: \(\text{volume} = \frac{106.85 \, \text{g}}{1.024 \, \text{g/mL}} = 104.34 \, \text{mL} = 0.10434 \, \text{L}\).
04

Rearrange the Osmotic Pressure Formula

Rearrange the osmotic pressure formula to solve for \(n\), number of moles of solute: \(n = \frac{\pi V}{RT}\). Substitute the known values: \(n = \frac{4.61 \, \text{atm} \cdot 0.10434 \, \text{L}}{0.0821 \, \frac{L·atm}{K·mol} \cdot 293.15 \, \text{K}}\).
05

Calculate the Number of Moles of Solute

Perform the calculation: \[n = \frac{4.61 \times 0.10434}{0.0821 \times 293.15} = 0.0197 \, \text{mol}\].
06

Determine the Molar Mass

Finally, use the relationship \(\text{molar mass} = \frac{\text{mass of solute}}{\text{moles of solute}}\):\[\text{molar mass} = \frac{6.85 \, \text{g}}{0.0197 \, \text{mol}} = 347.21 \, \text{g/mol}\].

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Key Concepts

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

Osmotic Pressure
Osmotic pressure, often represented by the symbol \(\pi\), is a crucial concept in chemistry, particularly when dealing with solutions. It refers to the pressure required to prevent the inward flow of water across a semipermeable membrane. This property is especially important in biological systems and various industrial applications.

The formula used to calculate osmotic pressure is given by:
  • \(\pi = \frac{n}{V}RT\)
where \(n\) stands for the number of moles of solute, \(V\) is the volume of the solution in liters, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.

Osmotic pressure helps us understand how solute concentration affects fluid movement between compartments or solutions. In the provided exercise, knowing the osmotic pressure allows us to use other known values, like temperature and the volume of the solution, to find the molar mass of the solute. It essentially plays a key role in identifying vital characteristics of the solution, such as concentration and molecular composition.
Ideal Gas Constant
The ideal gas constant, denoted as \(R\), is a significant factor in various chemical equations, including the osmotic pressure formula. This constant links energy quantities across the macroscopic and microscopic levels in different scenarios, providing a unified understanding of behaviors in gases and solutions.

Its standard value, in the context of osmotic pressure calculations, is:
  • \(R = 0.0821 \frac{L \cdot atm}{K \cdot mol}\)
Utilizing the ideal gas constant in formulas showcases its flexibility and importance in different chemical contexts. It ensures that when temperature, pressure, volume, and moles interact, the relationships are consistent and predictable. This reliability allows us to predict the behavior of gases and amounts of solvents or solutes required in solutions.

In calculating osmotic pressure and eventually molar mass, \(R\) acts as a crucial conversion factor that simplifies these interactions and links the physical properties involved in the equation.
Density Formula
The density formula is an indispensable tool in calculating the volume of substances, which further aids in understanding their concentration within a given space. Density is defined as mass per unit volume and is expressed as:
  • \( \text{Density} = \frac{\text{mass}}{\text{volume}} \)
Rearranging the formula helps us solve for the volume of a solution when the mass and density are known. This rearrangement is shown as:
  • \( \text{Volume} = \frac{\text{mass}}{\text{Density}} \)
In our given problem, the mass of the solution and its density were specified. This allowed us to determine the volume in liters, a critical step in calculating osmotic pressure and later, the molar mass of the solute.

Understanding the density formula and its application is fundamental since it connects mass with how spread out or concentrated a substance can be within a space, thereby impacting other calculations like molarity or osmotic pressure.
Temperature Conversion
Temperature conversion, particularly from Celsius to Kelvin, is essential in performing accurate chemical calculations. This conversion ensures uniformity and consistency when plugging temperature values into scientific formulas, where Kelvin is the preferred unit.

To convert Celsius to Kelvin, one must add:
  • 273.15 to the Celsius value
For example, in the exercise, the given temperature was \(20.0^{\circ}C\). In Kelvin, this is:
  • \(20.0 + 273.15 = 293.15 \text{ K}\)
Using Kelvin in scientific calculations maintains accuracy and prevents discrepancies, since Kelvin starts at absolute zero, aligning with the concept of energy measurement at the microscopic level.

Accurate temperature conversion facilitates correct computation of osmotic pressure, which directly depends on this precise value, aiding in calculations that ultimately lead to determining the properties of the solution, such as molar mass.

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Most popular questions from this chapter

The concentrated sulfuric acid we use in the laboratory is 98.0 percent \(\mathrm{H}_{2} \mathrm{SO}_{4}\) by mass. Calculate the molality and molarity of the acid solution. The density of the solution is \(1.83 \mathrm{~g} / \mathrm{mL}\).

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