(a) A chemist prepared a solution by dissolving \(1.230 \mathrm{~g}\) of \(\mathrm{KCl}\) in enough water to make \(150.0 \mathrm{~mL}\) of solution. What molar concentration of potassium chloride should appear on the label? (b) If the chemist had mistakenly used a 500 -mL. volumetric flask instead of the 150.0-mL flask in (a), what molar concentration of potassium chloride has the chemist actually prepared?

Short Answer

Expert verified
The correct molar concentration for the 150.0 mL solution is 0.109 M KCl, and for the 500 mL solution, it is 0.033 M KCl.

Step by step solution

01

Calculate the moles of KCl

Use the molar mass of KCl (74.55 g/mol) to convert the mass of KCl into moles using the formula: moles of KCl = mass of KCl / molar mass of KCl. For 1.230 g of KCl, the calculation is moles of KCl = 1.230 g / 74.55 g/mol.
02

Convert volume to liters

The volume should be in liters for molarity calculations. Since the volume is given in milliliters, convert it to liters by dividing by 1000. For 150.0 mL, the volume in liters is 0.1500 L.
03

Calculate molar concentration for 150.0 mL solution

Use the formula for molarity M = moles of solute / liters of solution. Plug in the values obtained from previous steps to calculate the molarity.
04

Calculate moles of KCl for the 500 mL solution

The amount of KCl doesn't change, only the volume of the solution does. The moles of KCl remain the same as calculated in Step 1.
05

Convert the volume of 500-mL flask to liters

Just like before, we convert the volume of 500 mL to liters by dividing by 1000, resulting in 0.500 L.
06

Calculate molar concentration for the 500 mL solution

Again, use the molarity formula M = moles of solute / liters of solution with the same number of moles from Step 1 but with the new volume converted to liters from Step 5.

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 Vaia!

Key Concepts

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

Molarity
Molarity is a measure of the concentration of a solute in a solution. It represents the number of moles of a solute that are dissolved in one liter of solution. The formula for calculating molarity is ewline \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \) ewline Using this formula, chemists can calculate how concentrated a solution is, which is important during the preparation of chemical solutions for various reactions or for titrations in volumetric analysis. When working with molarity, ensure that the volume is always in liters and the amount of solute is in moles to obtain accurate results. For example, if you dissolve a known mass of a substance in a specific volume of water, you can use the molar mass of the compound and the formula above to find its molarity.
Moles of Solute
The concept of 'moles of solute' is essential in chemistry, especially when involving molar concentration calculations. A mole is a unit that represents a set number of particles, atoms, ions, or molecules. In chemistry, it is commonplace to use Avogadro's number, which is approximately \(6.022 \times 10^{23}\), to represent one mole of any substance. ewline To calculate the moles of solute, use the formula: ewline \( \text{moles of solute} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \) ewline This transformation from mass to moles is critical, as it allows chemists to relate the mass of a substance to the actual number of particles contained within that mass, providing a way to quantify and work with substances at the molecular level.
Volumetric Analysis
Volumetric analysis, or titration, is a technique used to determine the concentration of a solution. It involves the careful addition of a standard solution, known as the titrant, to a solution of an analyte until the reaction between the two is complete. The point of completion is often indicated by a color change due to an indicator or by reaching an electrode potential in the case of potentiometric titrations. ewline Key to volumetric analysis is the use of volumetric flasks, which are precisely calibrated to contain a specific volume at a certain temperature. Accurate measurement of both the titrant and the analyte volume is paramount. If the volumes are not measured correctly, the resulting concentration calculations may be incorrect, as seen in the given exercise where using the incorrect flask volume led to a different molarity calculation.
Solution Concentration
Solution concentration refers to the amount of solute present in a given quantity of solvent or solution. Apart from molarity, there are other ways to express solution concentration, such as molality, normality, weight/volume percentage, and volume/volume percentage. Concentrations can also be communicated in terms of parts per million (ppm) or parts per billion (ppb), especially when dealing with very dilute solutions. ewline Understanding the concentration of a solution is crucial in fields such as chemistry, biology, medicine, and environmental science. It affects reaction rates, the outcome of titrations in volumetric analysis, and the effectiveness of solutions in applications such as intravenous fluids. Accurate determination and expression of solution concentration are necessary for replicating conditions in experiments and producing predictable results.

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

To prepare a very dilute solution, it is advisable to perform successive dilutions of a single prepared reagent solution, rather than to wcigh a very small mass or to measure a very small volume of stock chemical. A solution was prepared by transferring \(0.661 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to a \(250.0\)-mL volumetric flask and adding water to the mark. A \(1.000-\mathrm{mL}\). sample of this solution. was transferred to a 500 -mL. volumetric flask and diluted to the mark with water. Then \(10.0 \mathrm{~mL}\) of the diluted solution was transferred to a 250 -mL. flask and diluted to the mark with water. (a) What is the final concentration of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{\text {, in solution? (b) What }}\) mass of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is in this final solution? (The answer to the last question gives the amount that would have had to have been weighed out if the solution had been prepared directly.)

A chemist prepared an aqueous solution by mixing \(2.50 \mathrm{~g}\) of ammonium phosphate trihydrate, \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) and \(1.50 \mathrm{~g}\) of potassium phosphate, \(\mathrm{K}_{3} \mathrm{PO}_{4}\), with \(500 \mathrm{~g}\) of water. (a) Determine the number of moles of formula units of each compound that was measured. (b) How many moles of \(\mathrm{PO}_{4}{ }^{3-}\) are present in solution? (c) Calculate the mass of phosphate ions present in the solution. (d) What is the total mass of the water present in the solution?

Calculate the molality of the solute in each of the following solutions: (a) \(10.0 \mathrm{~g}\) of \(\mathrm{NaCl}\) dissolvcd in \(250 \mathrm{~g}\) water; (b) \(0.48 \mathrm{~mol}\) of \(\mathrm{KOH}\) dissolved in \(50.0 \mathrm{~g}\) water; (c) \(1.94 \mathrm{~g}\) of urea, \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\), dissolved in \(200 \mathrm{~g}\) water.

(a) What mass (in grams) of \(\mathrm{NaOH}\) must be mixed with \(250 \mathrm{~g}\) of water to prepare a \(0.22 \mathrm{~m} \mathrm{NaOH}\) solution? (b) Calculate the amount (in moles) of cthylene glycol, \(\mathrm{HOC}_{2} \mathrm{H}_{4} \mathrm{OH}\), that should be added to \(2.0 \mathrm{~kg}\) of water to prepare a \(0.44 \mathrm{~m} \mathrm{HOC}_{2} \mathrm{H}_{4} \mathrm{OH}\) (aq) solution. (c) Determine the amount (in moles) of HCl that must be dissolved in \(500 \mathrm{~g}\) water to prepare a \(0,0010 m\) HClaq) solution.

Calculate the mole fraction of each component in the following solutions: (a) \(2.5 .0 \mathrm{~g}\) of water and \(50.0 \mathrm{~g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\); (b) \(25.0 \mathrm{~g}\) of water and \(50.0 \mathrm{~g}\) of methanol, \(\mathrm{CH}_{3} \mathrm{OH}\); (c) a glucowe solution that is \(0.10 \mathrm{~m}\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)\).

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.

Sign-up for free