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Chapter 13: Problem 62

A solution of \(2.50 \mathrm{~g}\) of a compound of empirical formula \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{P}\) in \(25.0 \mathrm{~g}\) of benzene is observed to freeze at \(4.3^{\circ} \mathrm{C}\). Calculate the molar mass of the solute and its molecular formula.

Short Answer

Expert verified
The molar mass of the solute is 10.683 g/mol and the molecular formula is \(C_{6}H_{5}P\).

Step by step solution

01

Estimate Molecular Mass of Empirical Formula

Firstly, estimate the molecular mass of the empirical formula. According to the periodic table: C = 12.01 g/mol, H = 1.008 g/mol, and P = 30.97 g/mol. Thus, the molar mass of the empirical formula, \(C_{6}H_{5}P\), is: 6(12.01 g/mol) + 5(1.008 g/mol) + 30.97 g/mol = 108.122 g/mol.
02

Calculate Molality of the Solution

Afterwards, calculate the molality of the solution using the formula: molality (m) = moles of solute / mass of solvent (in kg). The moles of solute is given by the mass of solute divided by the molecular mass from the empirical formula: 2.50g / 108.122 g/mol = 0.0231 mol. The mass of benzene is 25.0 g, thus, in kg it is 0.025 kg. Therefore, the molality of the solution is 0.0231 mol / 0.025 kg = 0.924 m.
03

Find Freezing Point Depression of Benzene

Then, find the freezing point depression of benzene by subtracting the freezing point of the solution from the freezing point of pure benzene. The freezing point of pure benzene is 5.5°C. So, the freezing point depression, ΔTf, is: 5.5°C - 4.3°C = 1.2°C.
04

Calculate Molar Mass of Solute

Using the formula ΔTf = Kf x m, where Kf is the molal freezing point depression constant for benzene (5.12 °C/m), the molar mass of the solute can be calculated. Rearranging the formula gives: m = ΔTf / Kf = 1.2°C / 5.12 °C/m = 0.234 m. Therefore, molar mass of the solute = mass of the solute/moles of the solute = 2.5 g/0.234 m = 10.683 g/mol.
05

Determine the Molecular Formula

Calculate the ratio of the molar mass to the empirical formula mass: 10.683 g/mol / 108.122 g/mol = 0.099. Since this is approximately 1, the empirical and molecular formulas are the same, C6H5P.

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Most popular questions from this chapter

Consider two aqueous solutions, one of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) and the other of nitric acid \(\left(\mathrm{HNO}_{3}\right)\) both of which freeze at \(-1.5^{\circ} \mathrm{C}\). What other properties do these solutions have in common?

Give two exceptions to Henry's law.

Calculate the molalities of these aqueous solutions: (a) \(1.22 M\) sugar \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) solution (density of solution \(=1.12 \mathrm{~g} / \mathrm{mL}\) ), (b) \(0.87 \mathrm{M} \mathrm{NaOH}\) solution (density of solution \(=1.04 \mathrm{~g} / \mathrm{mL}),(\mathrm{c})\) \(5.24 \mathrm{M} \mathrm{NaHCO}_{3}\) solution (density of solution \(=\) \(1.19 \mathrm{~g} / \mathrm{mL})\)

Explain why ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is not soluble in cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right) .\)

Concentrated hydrochloric acid is usually available at 37.7 percent by mass. What is its concentration in molarity? (The density of the solution is \(1.19 \mathrm{~g} / \mathrm{mL}\).)
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