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Chapter 13: Problem 46

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

Short Answer

Expert verified
The molality of the KCl solution is \(3.88 \mathrm{~m}\). To make a \(0.12 \mathrm{~m}\) solution of sulfur in naphthalene, 4.684 g of sulfur must be dissolved in 100.0 g of naphthalene.

Step by step solution

01

Understand the definition of molality

Molality (m) is the number of moles of solute per kilogram of solvent. It is given by the formula: \(m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\)
02

Identify the given values

We are given: Moles of KCl = 1.12 mol Moles of water = 16.0 mol First, we need to convert moles of water to kilograms. We know that the molar mass of water is 18.015 g/mol.
03

Convert moles of water to kilograms

To convert moles of water to kilograms, we'll use the following equation: Kilograms of solvent = moles of solvent * molar mass of solvent * (1 kg/1000 g) Kilograms of water = 16.0 mol * 18.015 g/mol * (1 kg/1000 g) = 0.28824 kg
04

Calculate the molality of KCl solution

Using the molality formula from Step 1 and the values from Steps 2 and 3, we can now calculate the molality of KCl solution: \(m = \frac{1.12 \ \text{mol}}{0.28824 \ \text{kg}} = 3.88 \ \text{mol/kg}\) So, the molality of the KCl solution is 3.88 m. #b. Determine the mass of sulfur in a naphthalene solution#
05

Convert given molality to a mass-based formula

The given molality is 0.12 m. In this case, the molality represents the number of moles of sulfur \(\mathrm{(S_8)}\) per kilogram of naphthalene \(\mathrm{(C_{10}H_8)}\). We need to find the mass of sulfur required for a 100.0 g sample of naphthalene. Let's set up the mass-based equation: \(\text{molality} = \frac{\text{mass of sulfur} \cdot \text{conversion factor} }{\text{mass of naphthalene}}\) Here, the conversion factor is the molar mass of \(\mathrm{S_8}\).
06

Determine the molar mass of sulfur and naphthalene

The molar mass of sulfur \(\mathrm{(S_8)}\) is 256.48 g/mol. The molar mass of naphthalene \(\mathrm{(C_{10}H_8)}\) is 128.16 g/mol.
07

Calculate the mass of sulfur required

Using the equation from Step 1 and the values from Step 2, we can find the mass of sulfur required as follows: \(0.12 \ \text{m} = \frac{\text{mass of sulfur} \cdot 256.48 \ \text{g/mol} }{100.0 \ \text{g}}\) Now, we need to solve for the mass of sulfur: Mass of sulfur = \((0.12 \ \text{m} * 100.0 \ \text{g}) / 256.48 \ \text{g/mol} = 4.684 \ \text{g}\) So, 4.684 g of sulfur must be dissolved in 100.0 g of naphthalene to make a 0.12 m solution.

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Most popular questions from this chapter

Indicate whether each statement is true or false: \((\) a) \(\mathrm{NaCl}\) dissolves in water but not in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) because benzene is denser than water. (b) NaCl dissolves in water but not in benzene because water has a large dipole moment and benzene has zero dipole moment. (c) NaCl dissolves in water but not in benzene because the water-ion interactions are stronger than benzene-ion interactions.

(a) Calculate the vapor pressure of water above a solution prepared by adding \(22.5 \mathrm{~g}\) of lactose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\( to \)200.0 \mathrm{~g}\( of water at \)338 \mathrm{~K}$. (Vapor-pressure data for water are given in Appendix B.) (b) Calculate the mass of propylene glycol $\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{2}\right)\( that must be added to \)0.340 \mathrm{~kg}$ of water to reduce the vapor pressure by \(384 \mathrm{~Pa}\) at \(40^{\circ} \mathrm{C}\).

Compounds like sodium stearate, called "surfactants" in general, can form structures known as micelles in water, once the solution concentration reaches the value known as the critical micelle concentration (cmc). Micelles contain dozens to hundreds of molecules. The cmc depends on the substance, the solvent, and the temperature. At and above the \(\mathrm{cmc}\), the properties of the solution vary drastically. (a) The turbidity (the amount of light scattering) of solutions increases dramatically at the \(\mathrm{cmc}\). Suggest an explanation. (b) The ionic conductivity of the solution dramatically changes at the \(\mathrm{cmc}\). Suggest an explanation. (c) Chemists have developed fluorescent dyes that glow brightly only when the dye molecules are in a hydrophobic environment. Predict how the intensity of such fluorescence would relate to the concentration of sodium stearate as the sodium stearate concentration approaches and then increases past the \(\mathrm{cmc}\).

Benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) boils at $80.1^{\circ} \mathrm{C}\( and has a density of \)0.876 \mathrm{~g} / \mathrm{mL} .$ (a) When \(0.100 \mathrm{~mol}\) of a nondissociating solute is dissolved in $500 \mathrm{~mL}\( of \)\mathrm{C}_{6} \mathrm{H}_{6}$, the solution boils at \(79.52^{\circ} \mathrm{C}\). What is the molal boiling-point-elevation constant for \(\mathrm{C}_{6} \mathrm{H}_{6} ?\) (b) When \(10.0 \mathrm{~g}\) of a nondissociating unknown is dissolved in \(500 \mathrm{~mL}\) of $\mathrm{C}_{6} \mathrm{H}_{6}\(, the solution boils at \)79.23^{\circ} \mathrm{C}$. What is the molar mass of the unknown?

Consider water and glycerol, $\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}$. (a) Would you expect them to be miscible in all proportions? (b) List the intermolecular attractions that occur between a water molecule and a glycerol molecule.
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