/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 118 A solution is prepared by dissol... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Percent
Mass percent tells us how much of a solute is present in a solution in comparison to the total mass of the solution. It is expressed as a percentage and gives us a clear idea of the concentration by mass. To find the mass percent, use the formula:
  • Mass percent = \( \frac{\text{mass of solute}}{\text{mass of solution}} \times 100 \)
In our example, cesium chloride is the solute, and water is the solvent. By combining the mass of cesium chloride (52.3 g) and water (60.0 g) we find the mass of the solution. Plug these values into the formula above to determine the mass percent of cesium chloride in the solution. The mass percent indicates how concentrated the solution is, by showing the portion of cesium chloride relative to the total mass.
Molarity
Molarity measures the concentration of a solute in a solution relative to the volume. Specifically, it’s the number of moles of solute per liter of solution. Molarity is important when mixing chemicals because it tells you the exact concentration, allowing for consistency in experiments. To calculate molarity, apply the formula:
  • Molarity = \(\frac{\text{moles of solute}}{\text{volume of solution in liters}} \)
In the solution exercise, first convert the volume from milliliters to liters. With 63.3 mL given, this becomes 0.0633 L. Using the previously calculated moles of cesium chloride, use this formula to calculate the molarity. Molarity helps us understand how concentrated the cesium chloride is in every liter of the solution.
Molality
Molality is another way to express concentration, based instead on the mass of the solvent. Unlike molarity, which can change with temperature and pressure due to volume changes, molality is temperature-independent as it is based on mass. The formula for molality is:
  • Molality = \(\frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)
In the problem statement, convert the mass of the water from grams to kilograms. With 60 g of water present, that converts to 0.060 kg. With the moles of cesium chloride already known, calculate the molality using the formula above. Molality provides another measure of concentration, particularly useful in scenarios involving temperature variations.
Mole Fraction
Mole fraction is a dimensionless number that compares the number of moles of one component to the total number of moles in the solution. It provides insight into the composition of a solution without relying on mass or volume. To find the mole fraction, use the following formula:
  • Mole fraction = \(\frac{\text{moles of solute}}{\text{total moles in the solution}} \)
First, find the number of moles for each component—cesium chloride and water. Next, add them together to obtain the total moles. Use the moles of cesium chloride and the total moles to solve for the mole fraction. This value tells us how dominant cesium chloride is relative to water in the solution. Mole fraction is especially useful in thermodynamic calculations and applications involving phase changes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the solubility of \(\mathrm{O}_{2}\) in water at a partial pressure of \(\mathrm{O}_{2}\) of 120 torr at \(25^{\circ} \mathrm{C}\) . The Henry's law constant for \(\mathrm{O}_{2}\) is \(1.3 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot\) atm for Henry's law in the form \(C=k P\) where \(C\) is the gas concentration \((\mathrm{mol} / \mathrm{L})\)

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 \(45^{\circ} \mathrm{C}\) ?

Using the phase diagram for water and Raoult’s law, explain why salt is spread on the roads in winter (even when it is below freezing).

The term proof is defined as twice the percent by volume of pure ethanol in solution. Thus, a solution that is 95\(\%\) (by volume) ethanol is 190 proof. What is the molarity of ethanol in a 92 proof ethanol-water solution? Assume the density of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) , is 0.79 \(\mathrm{g} / \mathrm{cm}^{3}\) and the density of water is 1.0 \(\mathrm{g} / \mathrm{cm}^{3}\) .

The freezing point of \(t\) -butanol is \(25.50^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(9.1^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\) Usually \(t\) -butanol absorbs water on exposure to air. If the freezing point of a 10.0 -g sample of \(t\) -butanol is \(24.59^{\circ} \mathrm{C},\) how many grams of water are present in the sample?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.