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Chapter 4: Problem 32

A solution of ethanol ( \({C}_{2} {H}_{5} {OH}\) ) in water is prepared by dissolving 75.0 \({mL}\) of ethanol (density \(=0.79 {g} / {cm}^{3} )\) in enough

Short Answer

Expert verified
The molality of ethanol in the solution is approximately 3.024 mol/kg.

Step by step solution

01

Calculate the mass of ethanol

Given the volume of ethanol (75.0 \({mL}\)) and the density of ethanol (0.79 \({g}/{cm}^{3}\)), we can calculate the mass of ethanol. Since 1 \({mL}\) is equal to 1 \({cm}^{3}\), we can use density as follows: \( Mass\_of\_ethanol = Density \times Volume \) \( Mass\_of\_ethanol = 0.79 \frac{g}{cm^{3}} \times 75.0 cm^{3}\) Upon calculating, we find: \( Mass\_of\_ethanol = 59.25 g \)
02

Calculate the moles of ethanol

Now that we have calculated the mass of ethanol, we need to convert it into moles using the molar mass of ethanol, which is 12.01 (C) + 6*1.01 (H) + 16.00 (O) = 46.07 g/mol. \( Moles\_of\_ethanol = \frac{Mass\_of\_ethanol}{Molar\_mass} \) \( Moles\_of\_ethanol = \frac{59.25 g}{46.07 \frac{g}{mol}} \) Upon calculating, we find: \( Moles\_of\_ethanol = 1.286 mol \)
03

Calculate the mass of the solvent (water)

Knowing the volume of the solution (500.0 \({mL}\)) and the volume of ethanol (75.0 \({mL}\)), we can find the volume of water: \( Volume\_of\_water = Volume\_of\_solution - Volume\_of\_ethanol \) \( Volume\_of\_water = 500.0 mL - 75.0 mL \) \( Volume\_of\_water = 425 mL \) Next, we use the density of water (1.00 \({g}/{cm}^{3}\)) to calculate the mass of water: \( Mass\_of\_water = Density \times Volume\_of\_water \) \( Mass\_of\_water = 1.00 \frac{g}{cm^{3}} \times 425 cm^{3}\) Upon calculating, we find: \( Mass\_of\_water = 425 g \)
04

Calculate the molality of the solution

Molality is defined as the number of moles of solute (ethanol) per kilogram of solvent (water). We can calculate molality using the moles of ethanol and mass of water from the previous steps: \( Molality = \frac{Moles\_of\_ethanol}{Mass\_of\_water (in\, kg)} \) \( Molality = \frac{1.286 mol}{0.425 kg} \) Upon calculating, we find: \( Molality = 3.024 mol/kg \) Hence, the molality of ethanol in the solution is approximately 3.024 mol/kg.

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Most popular questions from this chapter

Carminic acid, a naturally occurring red pigment extracted from the cochineal insect, contains only carbon, hydrogen, and oxygen. It was commonly used as a dye in the first half of the nineteenth century. It is 53.66% C and 4.09% H by mass. A titration required 18.02 mL of 0.0406 M NaOH to neutralize 0.3602 g carminic acid. Assuming that there is only one acidic hydrogen per molecule, what is the molecular formula of carminic acid?

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