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Chapter 11: Problem 78

A solution contains 3.75 g of a nonvolatile pure hydrocarbon in 95 g acetone. The boiling points of pure acetone and the solution are $55.95^{\circ} \mathrm{C}\( and \)56.50^{\circ} \mathrm{C},$ respectively. The molal boiling- point constant of acetone is $1.71^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}$ . What is the molar mass of the hydrocarbon?

Short Answer

Expert verified
The molar mass of the hydrocarbon is approximately 122.92 g/mol. To find this, we used the boiling-point elevation formula, calculated the change in boiling point (ΔT), determined the molality of the solution, calculated the moles of solute (hydrocarbon), and finally divided the mass of the hydrocarbon by the moles of the hydrocarbon.

Step by step solution

01

Understand the boiling-point elevation formula

The formula for boiling-point elevation is: ΔT = Kb * molality Where ΔT is the change in boiling point due to the presence of solute, Kb is the molal boiling-point constant of the solvent, and molality is the molality of the solution (mol solute per kg of solvent). We are given the boiling points of the pure acetone and the solution, and the molal boiling-point constant (Kb) of acetone. With this information, we will be able to calculate the molality of the solution.
02

Calculate the change in boiling point (ΔT)

To calculate ΔT, we subtract the boiling point of pure acetone from the boiling point of the solution: ΔT = Boiling point of the solution - Boiling point of pure acetone ΔT = 56.50 °C - 55.95 °C = 0.55 °C
03

Calculate the molality of the solution

We use the boiling-point elevation formula to calculate the molality of the solution: Molality = ΔT / Kb Molality = 0.55 °C / 1.71 °C⋅kg/mol = 0.3216 mol/kg
04

Calculate the moles of solute (hydrocarbon)

As we have the molality of the solution and the amount of solvent in kg, we can calculate the moles of solute (hydrocarbon) present in the solution: Moles of solute = Molality * (mass of solvent in kg) Moles of solute = 0.3216 mol/kg * (95 g / 1000 g/kg) = 0.03052 mol
05

Calculate the molar mass of the hydrocarbon

Finally, we can calculate the molar mass of the hydrocarbon by dividing the mass of the hydrocarbon by the moles of the hydrocarbon: Molar mass = mass of hydrocarbon / moles of hydrocarbon Molar mass = 3.75 g / 0.03052 mol ≈ 122.92 g/mol The molar mass of the hydrocarbon is approximately 122.92 g/mol.

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Most popular questions from this chapter

A solution is prepared by mixing 50.0 \(\mathrm{mL}\) toluene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\right.\) $d=0.867 \mathrm{g} / \mathrm{cm}^{3} )\( with 125 \)\mathrm{mL}$ benzene $\left(\mathrm{C}_{6} \mathrm{H}_{6}, d=0.874 \mathrm{g} / \mathrm{cm}^{3}\right)$ Assuming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the toluene.

Which solvent, water or hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right),\) would you choose to dissolve each of the following? a. \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) b. \(\mathrm{CS}_{2}\) c. \(\mathrm{CH}_{3} \mathrm{OH}\) d. $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{CH}_{2} \mathrm{OH}$ e. \(\mathrm{HCl}\) f. \(\mathrm{C}_{6} \mathrm{H}_{6}\)

Formic acid \(\left(\mathrm{HCO}_{2} \mathrm{H}\right)\) is a monoprotic acid that ionizes only partially in aqueous solutions. A \(0.10-M\) formic acid solution is 4.2\(\%\) ionized. Assuming that the molarity and molality of the solution are the same, calculate the freezing point and the boiling point of 0.10\(M\) formic acid.

a. Calculate the freezing-point depression and osmotic pressure at $25^{\circ} \mathrm{C}\( of an aqueous solution containing 1.0 \)\mathrm{g} / \mathrm{L}$ of a protein (molar mass \(=9.0 \times 10^{4} \mathrm{g} / \mathrm{mol} )\) if the density of the solution is 1.0 \(\mathrm{g} / \mathrm{cm}^{3}\) b. Considering your answer to part a, which colligative property, freezing- point depression or osmotic pressure, would be better used to determine the molar masses of large molecules? Explain.

What volume of ethylene glycol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right),$ a nonelectrolyte,must be added to 15.0 \(\mathrm{L}\) water to produce an antifreeze solution with a freezing point of \(-25.0^{\circ} \mathrm{C} ?\) What is the boiling point of this solution? (The density of ethylene glycol is \(1.11 \mathrm{g} / \mathrm{cm}^{3},\) and the density of water is 1.00 \(\mathrm{g} / \mathrm{cm}^{3} . )\)
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