Chapter 2: Problem 50
' \(x\) ' grams of water is mixed in \(69 \mathrm{~g}\) of ethanol Mole fraction of ethanol in the resultant solution is \(0.6\). what is the value of ' \(\mathrm{x}\) ' in grams? (a) 54 (b) 36 (c) 180 (d) 18
Short Answer
Expert verified
The value of \(x\) in grams is 46.
Step by step solution
01
Understand Mole Fraction
Mole fraction of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components in the solution. It is represented by the symbol \(x\text{(component)}\). The mole fraction of ethanol is given as 0.6.
02
Calculate Moles of Ethanol
First, calculate the moles of ethanol using its mass and molar mass. The molar mass of ethanol \(C_2H_5OH\) is approximately \(46 \text{g/mol}\). Using the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}}\), calculate the moles of ethanol. \( \text{moles of ethanol} = \frac{69 \text{g}}{46 \text{g/mol}}\).
03
Relate Moles of Water and Ethanol
Let the moles of water be \(n_{H_2O}\). The mole fraction equation can be defined as \(0.6 = \frac{\text{moles of ethanol}}{\text{moles of ethanol} + n_{H_2O}}\).
04
Solve for Moles of Water
Plug in the value for moles of ethanol into the equation from Step 3 and solve for the moles of water, \(n_{H_2O}\).
05
Convert Moles of Water to Grams
Convert the moles of water to grams by using the molar mass of water (approximately \(18 \text{g/mol}\)): \(x = n_{H_2O} \times \text{molar mass of water}\ \ (18 \text{g/mol})\).
06
Calculate the Value of \(x\) in Grams
Substitute the calculated value of \(n_{H_2O}\) from Step 4 into the equation \(x = n_{H_2O} \times 18 \text{g/mol}\) to find the mass of water in grams.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Moles Calculation
The concept of moles is fundamental in chemistry, acting as a bridge between the mass of a substance and the number of particles it contains. A mole is defined as the amount of substance that contains as many entities (atoms, molecules, ions, or other particles) as there are atoms in 12 grams of carbon-12.
To calculate moles, one typically uses the formula:
\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
For instance, to find the moles of ethanol required in our textbook exercise, you divide the given mass of ethanol (69 g) by its molar mass (46 g/mol). This approach helps to quantify the amount of a substance involved in chemical reactions and is crucial for understanding solution composition, which in turn affects physical properties like boiling point and vapor pressure.
To calculate moles, one typically uses the formula:
\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
For instance, to find the moles of ethanol required in our textbook exercise, you divide the given mass of ethanol (69 g) by its molar mass (46 g/mol). This approach helps to quantify the amount of a substance involved in chemical reactions and is crucial for understanding solution composition, which in turn affects physical properties like boiling point and vapor pressure.
Solution Composition
Understanding the composition of a solution is key to many areas in chemistry, especially when it comes to preparing solutions with precise concentrations. Solution composition can be described in various ways, with mole fraction being one of them. The mole fraction, denoted as \( x \), is the ratio of the number of moles of a particular component to the total number of moles in the mixture.
In our exercise, the mole fraction of ethanol is given as 0.6. This means that for every one mole of the solution, 0.6 moles are ethanol. To fully understand the composition, one would need to calculate the mole fraction of water as well, which can be deduced from the mole fraction of ethanol since the sum of the mole fractions of all components in a solution always equals one. The mole fraction provides an insight into the relative quantities of each component and is particularly useful because it is temperature-independent, unlike other concentration units.
In our exercise, the mole fraction of ethanol is given as 0.6. This means that for every one mole of the solution, 0.6 moles are ethanol. To fully understand the composition, one would need to calculate the mole fraction of water as well, which can be deduced from the mole fraction of ethanol since the sum of the mole fractions of all components in a solution always equals one. The mole fraction provides an insight into the relative quantities of each component and is particularly useful because it is temperature-independent, unlike other concentration units.
Chemical Molar Mass
Chemical molar mass, often simply referred to as molar mass, is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. The units typically used are grams per mole (g/mol), and it reflects how much one mole of a substance weighs.
To find the molar mass of a compound, like ethanol (\(C_2H_5OH\)), you need to sum the molar masses of all the atoms in the molecule. Carbon has a molar mass of approximately 12.01 g/mol, hydrogen about 1.008 g/mol, and oxygen about 16.00 g/mol. This calculation is essential when converting between moles and grams, helping us to understand the amount of a substance on a molecular level. For example, in the problem given, the molar mass of ethanol allowed us to calculate its moles, a step that was vital to calculate the resultant mole fraction of the solution.
To find the molar mass of a compound, like ethanol (\(C_2H_5OH\)), you need to sum the molar masses of all the atoms in the molecule. Carbon has a molar mass of approximately 12.01 g/mol, hydrogen about 1.008 g/mol, and oxygen about 16.00 g/mol. This calculation is essential when converting between moles and grams, helping us to understand the amount of a substance on a molecular level. For example, in the problem given, the molar mass of ethanol allowed us to calculate its moles, a step that was vital to calculate the resultant mole fraction of the solution.