A solution is made containing 20.8 g of phenol (C 6 H 5 OH) in 425 g of ethanol (CH 3 CH 2 OH). Calculate (a) the mole fraction of phenol,

well, everyone's on this video of being asked what is the mole fraction of the components of a solution that's made by 2.15 g of Ben's OIC acid dissolved in 50 g of acetone were also given the molar masses of each component here. So first we can go ahead and start by calculating for the moles of each component. Starting off with our benzoate acid. So given the mass year of 2.15 g of Ben's weak acid, we can go ahead and use the molar mass to convert our grams into moles. So for every 122.12 g we have one mole of Ben's OIC acid. So you see here that the units of grams will cancel. And finally playing this into my calculator, I get the value of 0. units being moles. And same thing with our acetone. So given that we have 50 g of acetone again, we'll use the molar mass here to convert our grams into moles. So for every 58.8 g we have one mole of a stone. You see here that the gram unit will cancel. Now putting this into my calculator, I get the value of 0. moles. And with this information for the molds of each component, we can calculate for the mole fraction of each. So starting off with the mole fraction of Ben's weak acid. This is equal to well the moles of Arbenz folic acid on our numerator numerator. So that 0.176056 moles. And the denominator has to be the total amount of moles. So it's going to add the moles of both, that 0.176056 plus 0.8608815. And then once you get that, we get the value of 0.02 mol being the mole fraction of Urban's OIC acid, calculating for the mole fraction of acetone. Again, As our numerator, we need the moles of our stone. So that's 0. 08815 moles. Denominator will be exact same. So that's 0.176056 plus 0.8608815. So once you put that into a calculator, you get the value of 0.98. So, these values here are going to be my final answer for my problem. All right. Thank you all so much for watching

Ch.13 - Properties of Solutions

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Brown 15th Edition

Ch.13 - Properties of Solutions

Problem 42a

Chapter 13, Problem 42a

A solution is made containing 20.8 g of phenol (C₆H₅OH) in 425 g of ethanol (CH₃CH₂OH). Calculate (a) the mole fraction of phenol,

Verified step by step guidance

First, calculate the number of moles of phenol (C6H5OH). Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). The molar mass of phenol is approximately 94.11 g/mol.

Next, calculate the number of moles of ethanol (CH3CH2OH). Use the same formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). The molar mass of ethanol is approximately 46.07 g/mol.

Add the moles of phenol and the moles of ethanol to find the total number of moles in the solution.

Calculate the mole fraction of phenol using the formula: \( \text{mole fraction of phenol} = \frac{\text{moles of phenol}}{\text{total moles in solution}} \).

Ensure that the mole fraction is a dimensionless number, typically expressed as a decimal.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Mole Fraction

Mole fraction is a way of expressing the concentration of a component in a mixture. It is calculated by dividing the number of moles of the component by the total number of moles of all components in the solution. This dimensionless quantity provides insight into the relative amounts of substances in a mixture, which is particularly useful in solutions.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a molecule. Knowing the molar mass is essential for converting between grams and moles, which is necessary for calculating mole fractions in a solution.

Solution Composition

Solution composition refers to the relative amounts of solute and solvent in a solution. In this case, phenol is the solute and ethanol is the solvent. Understanding the composition is crucial for calculations involving concentrations, such as mole fraction, as it helps determine how much of each component is present in the solution.