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Chapter 13: Problem 99

A "canned heat" product used to warm chafing dishes consists of a homogeneous mixture of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and paraffin that has an average formula of \(\mathrm{C}_{24} \mathrm{H}_{50}\). What mass of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) should be added to \(620 \mathrm{~kg}\) of the paraffin in formulating the mixture if the vapor pressure of ethanol at \(35^{\circ} \mathrm{C}\) over the mixture is to be 8 torr? The vapor pressure of pure ethanol at \(35^{\circ} \mathrm{C}\) is 100 torr.

Short Answer

Expert verified
Approximately 6.9 kg of ethanol (C2H5OH) should be added to 620 kg of paraffin to formulate the mixture with an 8 torr vapor pressure of ethanol at 35 °C.

Step by step solution

01

1. Recall Raoult's Law formula

Raoult's Law states: \[P_A = x_A P^*_A\] where \(P_A\) is the vapor pressure of component A over the mixture, \(x_A\) is the mole fraction of component A in the mixture, and \(P^*_A\) is the vapor pressure of the pure component A. In our case, A is ethanol (C2H5OH).
02

2. Rearrange the formula to find the mole fraction of ethanol

We need to find the mole fraction of ethanol in the mixture, so we'll rearrange Raoult's Law formula: \[x_A = \frac{P_A}{ P^*_A}\]
03

3. Calculate the mole fraction of ethanol

We are given the vapor pressure of ethanol over the mixture (\(P_A\)) to be 8 torr and the vapor pressure of pure ethanol (\(P^*_A\)) to be 100 torr. Substituting the values into the rearranged formula, we get: \[x_A = \frac{8 \, \text{torr}}{100 \, \text{torr}} = 0.08\]
04

4. Relate the mole fraction to mass and mole

We know that mole fraction (\(x_A\)) can be represented as \[x_A = \frac{\text{moles of } A}{\text{moles of } A + \text{moles of} \, B}\] where B is paraffin (\(C_{24}H_{50}\)).
05

5. Convert mass to moles

In order to use the mole fraction equation, we'll need to convert the mass of ethanol and paraffin into moles. We'll denote ethanol's mass as \(m_{C_2H_5OH}\) and paraffin's mass as \(m_{C_{24}H_{50}}\). We are given the mass of paraffin in the mixture (\(m_{C_{24}H_{50}}\)) as 620 kg, and we need to find the mass of ethanol (\(m_{C_2H_5OH}\)). The molecular weights are: \[M_{C_2H_5OH} = 46 \, \frac{g}{\text{mole}}\] and \[M_{C_{24}H_{50}} = 338 \, \frac{g}{\text{mole}}\] Converting the mass of paraffin to grams and moles, we get: \[n_{C_{24}H_{50}} = \frac{620 \, \text{kg}}{1} \times \frac{1000 \, g}{1 \, \text{kg}} \times \frac{1 \, \text{mole}}{338 \, g} = 1834.32 \, \text{moles}\]
06

6. Calculate the moles of ethanol

Using the mole fraction equation from step 4 and the mole fraction of ethanol from step 3, we can now solve for the moles of ethanol: \[0.08 = \frac{n_{C_2H_5OH}}{n_{C_2H_5OH} + 1834.32}\] Rearranging for \(n_{C_2H_5OH}\) and solving, we get: \[n_{C_2H_5OH} = 149.6 \, \text{moles}\]
07

7. Convert moles of ethanol to mass

Using the molecular weight of ethanol (\(M_{C_2H_5OH}\)) from step 5, we can convert the number of moles of ethanol back to mass: \[m_{C_2H_5OH} = 149.6 \, \text{moles} \times \frac{46\, g}{1\, \text{mole}} = 6881.6 \, g\]
08

8. Conclusion

6881.6 g, or approximately 6.9 kg of ethanol (C2H5OH), should be added to 620 kg of paraffin to formulate the mixture with an 8 torr vapor pressure of ethanol at 35 °C.

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Most popular questions from this chapter

At ordinary body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the solubility of \(\mathrm{N}_{2}\) in water in contact with air at ordinary atmospheric pressure \((1.0 \mathrm{~atm})\) is \(0.015 \mathrm{~g} / \mathrm{L}\). Air is approximately \(78 \mathrm{~mol} \% \mathrm{~N}_{2}\). Calculate the number of moles of \(\mathrm{N}_{2}\) dissolved per liter of blood, which is essentially an aqueous solution. At a depth of \(100 \mathrm{ft}\) in water, the pressure is \(4.0 \mathrm{~atm}\). What is the solubility of \(\mathrm{N}_{2}\) from air in blood at this pressure? If a scuba diver suddenly surfaces from this depth, how many milliliters of \(\mathrm{N}_{2}\) gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

Commercial aqueous nitric acid has a density of \(1.42 \mathrm{~g} / \mathrm{mL}\) and is 16 M. Calculate the percent \(\mathrm{HNO}_{3}\) by mass in the solution.

Adrenaline is the hormone that triggers the release of extra glucose molecules in times of stress or emergency. A solution of \(0.64 \mathrm{~g}\) of adrenaline in \(36.0 \mathrm{~g}\) of \(\mathrm{CCl}_{4}\) elevates the boiling point by \(0.49^{\circ} \mathrm{C}\). Is the molar mass of adrenaline calculated from the boiling-point elevation in agreement with the following structural formula?

(a) What is the molality of a solution formed by dissolving \(1.12 \mathrm{~mol}\) of \(\mathrm{KCl}\) in \(16.0 \mathrm{~mol}\) of water? (b) How many grams of sulfur \(\left(\mathrm{S}_{8}\right)\) must be dissolved in \(100.0 \mathrm{~g}\) naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) to make a \(0.12 \mathrm{~m}\) solution?

(a) A sample of hydrogen gas is generated in a closed container by reacting \(2.050 \mathrm{~g}\) of zinc metal with \(15.0 \mathrm{~mL}\) of 1.00 \(M\) sulfuric acid. Write the balanced equation for the reaction, and calculate the number of moles of hydrogen formed, assuming that the reaction is complete. (b) The volume over the solution is \(122 \mathrm{~mL}\). Calculate the partial pressure of the hydrogen gas in this volume at \(25^{\circ} \mathrm{C}\), ignoring any solubility of the gas in the solution. (c) The Henry's law constant for hydrogen in water at \(25^{\circ} \mathrm{C}\) is \(7.8 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) -atm. Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hydrogen in part (b)?
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