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Chapter 11: Problem 72

A solution contains \(3.75 \mathrm{~g}\) of a nonvolatile pure hydrocarbon in \(95 \mathrm{~g}\) 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 boilingpoint 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 122.56 g/mol.

Step by step solution

01

To calculate the change in boiling point (ΔTb), subtract the boiling point of the pure solvent from the boiling point of the solution: ΔTb = Boiling Point of Solution - Boiling Point of Pure Acetone ΔTb = \(56.50^{\circ} \mathrm{C}\) - \(55.95^{\circ} \mathrm{C}\) ΔTb = \(0.55^{\circ} \mathrm{C}\) #Step 2: Calculate Molality (m) Using Boiling Point Constant (Kb)#

We know that ΔTb = Kb * molality. Therefore, we can find the molality (m) by dividing the change in boiling point by the boiling point constant: molality = ΔTb / Kb molality = \(0.55^{\circ} \mathrm{C}\) / \(1.71^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\) molality = 0.322 mol/kg #Step 3: Calculate Moles of Hydrocarbon (n)#
02

Now that we have the molality, we can calculate the number of moles (n) of hydrocarbon in 1 kg of acetone using the formula: n = molality * mass of solvent (in kg) n = 0.322 mol/kg * (95 g / 1000 g/kg) # Convert mass of acetone to kg n = 0.03061 mol #Step 4: Calculate Molar Mass of Hydrocarbon (M)#

Finally, we can find the molar mass (M) of the hydrocarbon by dividing the mass of hydrocarbon (given) by the number of moles (calculated in Step 3): M = mass of hydrocarbon / moles of hydrocarbon M = 3.75 g / 0.03061 mol M ≈ 122.56 g/mol The molar mass of the hydrocarbon is approximately 122.56 g/mol.

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Most popular questions from this chapter

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 \(-25.0^{\circ} \mathrm{C} ?\) 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 \(1.00 \mathrm{~g} / \mathrm{cm}^{3} .\) )

A \(2.00-\mathrm{g}\) sample of a large biomolecule was dissolved in \(15.0 \mathrm{~g}\) carbon tetrachloride. The boiling point of this solution was determined to be \(77.85^{\circ} \mathrm{C}\). Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling-point constant is \(5.03^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\), and the boiling point of pure carbon tetrachloride is \(76.50^{\circ} \mathrm{C}\).

What mass of glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)\), a nonelectrolyte, must be dissolved in \(200.0 \mathrm{~g}\) water to give a solution with a freezing point of \(-1.50^{\circ} \mathrm{C} ?\)

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A solution of sodium chloride in water has a vapor pressure of \(19.6\) torr at \(25^{\circ} \mathrm{C}\). What is the mole fraction of solute particles in this solution? What would be the vapor pressure of this solution at \(45^{\circ} \mathrm{C} ?\) The vapor pressure of pure water is \(23.8\) torr at \(25^{\circ} \mathrm{C}\) and \(71.9\) torr at \(45^{\circ} \mathrm{C}\) and assume sodium chloride exists as \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions in solution.
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