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Chapter 11: Problem 118

A solution is prepared by dissolving 52.3 g cesium chloride in 60.0 g water. The volume of the solution is 63.3 \(\mathrm{mL}\) . Calculate the mass percent, molarity, molality, and mole fraction of the CsCl solution.

Short Answer

Expert verified
The mass percent of cesium chloride (CsCl) in the solution is 46.5%. The molarity of the solution is 12.42 M, and the molality is 13.16 m. The mole fraction of CsCl in the solution is 0.0501.

Step by step solution

01

Calculate the moles of CsCl

To calculate the moles of CsCl, we will use its molar mass: CsCl = 132.9 g/mol. moles of CsCl = (mass of CsCl) / (molar mass of CsCl) = \(\frac{52.3 \ \mathrm{g}}{132.9 \ \mathrm{g/mol}}\) Calculate the moles of CsCl using the equation above.
02

Calculate the moles of water

To calculate the moles of water, we use its molar mass: H2O = 18.015 g/mol. moles of H2O = (mass of H2O) / (molar mass of H2O) = \(\frac{60.0 \ \mathrm{g}}{18.015 \ \mathrm{g/mol}}\) Calculate the moles of water using the equation above.
03

Calculate mass percent

Mass percent = \(\frac{\text{mass of solute}}{\text{mass of solution}} \times 100\) Mass of the solution = mass of CsCl + mass of water = 52.3 g + 60.0 g Calculate the mass percent of CsCl using the mass of solute and solution.
04

Calculate molarity

Molarity = \(\frac{\text{moles of solute}}{\text{volume of the solution in liters}}\) First, convert the volume of the solution from mL to liters: 63.3 mL × \(\frac{1 \ \mathrm{L}}{1000 \ \mathrm{mL}}\) = 0.0633 L Next, calculate the molarity of the solution using the moles of CsCl and volume of the solution in liters.
05

Calculate molality

Molality = \(\frac{\text{moles of solute}}{\text{mass of solvent in kg}}\) First, convert the mass of water (the solvent) from grams to kilograms: 60 g × \(\frac{1 \ \mathrm{kg}}{1000 \ \mathrm{g}}\) Next, calculate the molality of the solution using the moles of CsCl and the mass of water (solvent) in kilograms.
06

Calculate mole fraction

Mole fraction = \(\frac{\text{moles of solute}}{\text{total moles in the solution}}\) Total moles = moles of CsCl + moles of H2O Calculate the mole fraction of CsCl using the moles of CsCl and total moles in the solution.

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Most popular questions from this chapter

Liquid A has vapor pressure \(x\) , and liquid \(\mathrm{B}\) has vapor pressure \(y .\) What is the mole fraction of the liquid mixture if the vapor above the solution is \(30 . \% \mathrm{A}\) by moles? $50 . \% \mathrm{A} ? 80 . \% \mathrm{A}\( ? (Calculate in terms of \)x\( and \)y . )$ Liquid A has vapor pressure \(x,\) liquid \(\mathrm{B}\) has vapor pressure y. What is the mole fraction of the vapor above the solution if the liquid mixture is \(30 . \% \mathrm{A}\) by moles? $50 . \% \mathrm{A} ? 80 . \% \mathrm{A}\( ? (Calculate in terms of \)x\( and \)y . )$

For each of the following, choose the pair of substances you would expect to give the most ideal solution. Explain your choices. a. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CH}_{3}\) and $\mathrm{CH}_{3}\left(\mathrm{CH}_{25} \mathrm{CH}_{3} \text { or } \mathrm{H}_{2} \mathrm{O} \text { and }\right.$ \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{3}\) b. \(\mathrm{CH}_{3} \mathrm{OH}\) and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\( or \)\mathrm{CH}_{3} \mathrm{F}\( and \)\mathrm{HF}$

An aqueous solution containing glucose has a vapor pressure of 19.6 torr at \(25^{\circ} \mathrm{C}\). 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 \(45^{\circ} \mathrm{C} .\) If the glucose in the solution were substituted with an equivalent amount (moles) of \(\mathrm{NaCl}\), what would be the vapor pressure at \(45^{\circ} \mathrm{C}\) ?

Rationalize the trend in water solubility for the following simple alcohols:

For each of the following pairs, predict which substance is more soluble in water. a. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) or \(\mathrm{NH}_{3}\) b. \(\mathrm{CH}_{3} \mathrm{CN}\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) or $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$ d. \(\mathrm{CH}_{3} \mathrm{OH}\) or $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ e. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\) f. \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) or $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$
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