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Chapter 3: Problem 29

A mixture of methanol (methyl alcohol) and water contains \(60.0 \%\) water by mass. (a) Assuming volume additivity of the components, estimate the specific gravity of the mixture at \(20^{\circ} \mathrm{C} .\) What volume (in liters) of this mixture is required to provide 150 mol of methanol? (b) Repeat Part (a) with the additional information that the specific gravity of the mixture at \(20^{\circ} \mathrm{C}\) is 0.9345 (making it unnecessary to assume volume additivity). What percentage error results from the volume- additivity assumption?

Short Answer

Expert verified
The specific gravity of the methanol-water mixture at \(20^{\circ}C\) is 0.905 (assuming volume additivity). The volume of the mixture required to provide 150 mol of methanol is approximately 6.063 L. If the specific gravity is given as 0.9345, then the calculated volume remains the same, but there is no error in the volume estimated using volume additivity assumption.

Step by step solution

01

(a) Step 1: Calculate the mass of methanol and water in the mixture

First calculate the mass of the methanol and the water in the mixture. Since we know the mixture is 60% water by mass, it must contain 40% methanol by mass. So, if we take the total mass of the mixture as 100 g, the mass of water = \(0.60 \times 100 g = 60 g\) and the mass of methanol = \(0.40 \times 100 g = 40 g\).
02

(a) Step 2: Calculate the volume of methanol and water in the mixture

Using known densities of methanol (\(0.792 g/cm^3\) or \(792 kg/m^3\)) and water (\(1 g/cm^3\) or \(1000 kg/m^3\)) at \(20^{\circ}C\), calculate the volumes of the components: Volume of water = \(\frac{60 g}{1 g/cm^3} = 60 cm^3\), Volume of methanol = \(\frac{40 g}{0.792 g/cm^3} = 50.5 cm^3\)
03

(a) Step 3: Calculate the total volume and density of the mixture

Assuming volume additivity, the total volume of the mixture = volume of water + volume of methanol = \(60 cm^3 + 50.5 cm^3 = 110.5 cm^3\). Then, calculate the density of the mixture: Density = \(\frac{total mass}{total volume}\) = \(\frac{100 g}{110.5 cm^3} = 0.905 g/cm^3\)
04

(a) Step 4: Calculate the specific gravity of the mixture

The specific gravity of the mixture (which is dimensionless) is the ratio of its density to the density of water. So, Specific gravity = \(\frac{density of the mixture}{density of water}\) = \(\frac{0.905 g/cm^3}{1 g/cm^3} = 0.905\)
05

(a) Step 5: Calculate the number of moles and volume of methanol in the mixture

From molar mass of methanol (32.04 g/mol), calculate the number of moles in 40 g: Number of moles = \(\frac{mass}{molar mass} = \frac{40g}{32.04 g/mol} = 1.248 mol\). We need 150 mol, so multiply the volume of methanol calculated in step 2 with the molar ratio. This gives: Volume = \(50.5 cm^3 \times \frac{150 mol}{1.248 mol}\) = \(6063 cm^3\) or 6.063 L
06

(b) Step 1: Calculate the density, total mass, and mass of methanol using given specific gravity

Using the given specific gravity (0.9345), calculate the density of the mixture: Density = Specific gravity x density of water = \(0.9345 g/cm^3 \times 1 g/cm^3 = 0.9345 g/cm^3\). With the total mass of the mixture = 100 g and the mass of the water = 60 g (assuming 60% water by mass, as given), it is calculated that the mass of the methanol = total mass - mass of water = \(100 g - 60 g = 40 g\).
07

(b) Step 2: Calculate the volume of methanol using known density

Calculate the volume of methanol using its known density at \(20^{\circ}C = 0.792 g/cm^3\) : Volume = \(\frac{mass}{density} = \frac{40 g}{0.792 g/cm^3} = 50.5 cm^3 \)
08

(b) Step 3: Calculate the required volume of methanol

Same as in (a) Step 5, calculate the required volume for 150 mol: Volume = \(50.5 cm^3 \times \frac{150 mol}{1.248 mol}\) = \(6063 cm^3\) or 6.063 L
09

(b) Step 4: Calculate the percentage error

The volume from volume-additivity assumption in part (a) equals the volume from given specific gravity in part (b). Thus, the percentage error = \(0\%\)

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Key Concepts

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

Specific Gravity Estimation
When it comes to understanding the properties of a mixture, one key aspect is determining its specific gravity. Specific gravity is a measure of density relative to the density of another reference substance, typically water at 4掳C (which has a density of 1 g/cm鲁). It's an important factor in various industries because it can influence how substances mix and react with each other.

For instance, in the problem at hand, we estimated the specific gravity of a methanol-water mixture by first calculating the masses of each component. Given that the density of the mixture equals the total mass divided by the total volume, and the specific gravity is the ratio of the mixture's density to that of water, we can deduce that specific gravity has no units. It is a useful quantity that can help predict the behavior of the mixture in different conditions, such as flotation and mixing processes.

Understanding specific gravity estimation in this scenario is crucial in real-world applications, such as designing chemical processes or the production of various consumer goods. It's also pertinent when dealing with environmental regulations regarding the discharge of industrial effluents, as the specific gravity of the discharged mixture can affect its dispersion in water bodies.
Mixture Composition Analysis
Mixture composition analysis is essential for understanding the properties and behavior of compounds within a given mixture. It involves breaking down a mixture into its individual components quantitatively. In our exercise, we analyzed the composition of a methanol-water mixture by mass percentage, which provided vital information to conduct further calculations such as the estimation of specific gravity. Composition analysis can be extended to include factors like molarity, molality, or normality of solutions, depending on the needs of the analysis.

By delving deep into mixture composition analysis, students and professionals can make precise adjustments to reaction conditions, optimize the product yield in industrial processes, or ensure the quality of pharmaceuticals. It's also a cornerstone in environmental science for assessing pollution levels and in the food industry for nutritional content calculation. Engaging with this type of analysis allows for a more controlled and predictive handling of chemical processes.
Molar Mass and Volume Calculations
Understanding molar mass and volume calculations is pivotal for chemists and chemical engineers alike. The molar mass of a substance is the weight of one mole of that compound and is expressed in grams per mole (g/mol). It enables us to convert between mass and moles, aiding in further calculations such as determining the volume of a substance needed for a reaction.

In the given problem, we've used molar mass to determine how much methanol is present in a given mass of mixture and then calculated the volume required to obtain 150 moles of methanol. These calculations are routine when preparing solutions for reactions, scaling up experiments to industrial production, or assessing the feasibility of reactions. Moreover, understanding the relationship between molar mass, the number of moles, and volume is crucial for titrations, stoichiometry, and preparation of standard solutions in analytical chemistry.

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