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Chapter 13: Problem 25

What is the molality of para-dichlorobenzene in a solution prepared by dissolving \(2.65 \mathrm{g} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}\) in \(50.0 \mathrm{mL}\) benzene \((d=0.879 \mathrm{g} / \mathrm{mL}) ?\)

Short Answer

Expert verified
The molality of the solution is \(0.410 \, mol/kg\).

Step by step solution

01

Calculate the molar mass of para-dichlorobenzene

The molar mass of para-dichlorobenzene (\(C_{6}H_{4}Cl_{2}\)) can be calculated as follows: \(6(12.01 \mathrm{g/mol}) + 4(1.008 \mathrm{g/mol}) + 2(35.45 \mathrm{g/mol}) = 146.98 \mathrm{g/mol}\).
02

Calculate the number of moles of para-dichlorobenzene

The number of moles of para-dichlorobenzene can be calculated by dividing the given mass by the molar mass: \((2.65 \mathrm{g}) / (146.98 \mathrm{g/mol}) = 0.0180 \mathrm{mol}\).
03

Convert the volume of benzene to mass

The volume of benzene can be converted to a mass by multiplying it by the given density: \((50.0 \mathrm{mL}) \times (0.879 \mathrm{g/mL}) = 43.95 \mathrm{g}\). Then convert the mass of benzene from grams to kilograms: \((43.95 \mathrm{g}) / (1000 \mathrm{g/kg}) = 0.04395 \mathrm{kg}\).
04

Calculate the molality of the solution

Molality is calculated by dividing the number of moles of the solute by the mass of the solvent in kilograms: \((0.0180 \mathrm{mol}) / (0.04395 \mathrm{kg}) = 0.410 \mathrm{mol/kg}\).

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Most popular questions from this chapter

What are the partial and total vapor pressures of a solution obtained by mixing 35.8 g benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\) and \(56.7 \mathrm{g}\) toluene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3},\) at \(25^{\circ} \mathrm{C} ? \mathrm{At} 25^{\circ} \mathrm{C}\) the vapor pressure of \(\mathrm{C}_{6} \mathrm{H}_{6}=95.1 \mathrm{mm} \mathrm{Hg} ;\) the vapor pressure of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}=28.4 \mathrm{mmHg}\).

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