Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A volume of \(35.2 \mathrm{~mL}\) of a \(1.66 \mathrm{M} \mathrm{KMnO}_{4}\) solution is mixed with \(16.7 \mathrm{~mL}\) of a \(0.892 \mathrm{M} \mathrm{KMnO}_{4}\) solution. Calculate the concentration of the final solution.

Short Answer

Expert verified
The final concentration is 1.413 M.

Step by step solution

01

Calculate moles in the first solution

First, determine the number of moles of \( \mathrm{KMnO}_{4} \) in the first solution using its volume and concentration. The formula to calculate moles is: \( \text{moles} = \text{concentration (M)} \times \text{volume (L)} \). Convert the volume from mL to L by dividing by 1000: \( 35.2 \, \text{mL} = 0.0352 \, \text{L} \). Then, calculate the moles, \( \text{moles} = 1.66 \, \text{M} \times 0.0352 \, \text{L} = 0.058432 \, \text{moles} \).
02

Calculate moles in the second solution

Next, calculate the moles of \( \mathrm{KMnO}_{4} \) in the second solution. The volume is \( 16.7 \, \text{mL} = 0.0167 \, \text{L} \). Using the formula for moles, \( \text{moles} = 0.892 \, \text{M} \times 0.0167 \, \text{L} = 0.0148964 \, \text{moles} \).
03

Calculate total moles

Add the moles from the two solutions to find the total moles of \( \mathrm{KMnO}_{4} \). \( 0.058432 \, \text{moles} + 0.0148964 \, \text{moles} = 0.0733284 \, \text{moles} \).
04

Calculate total volume of the mixture

To find the total volume of the mixture, sum the individual volumes of the solutions. In liters, \( 0.0352 \, \text{L} + 0.0167 \, \text{L} = 0.0519 \, \text{L} \).
05

Calculate the final concentration

Finally, use the total moles and the total volume to find the concentration of the final solution using the formula \( \text{concentration} = \frac{\text{moles}}{\text{volume (L)}} \). \( \frac{0.0733284 \, \text{moles}}{0.0519 \, \text{L}} = 1.413 \, \text{M} \).

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.

Understanding Molarity
Molarity is a crucial concept in chemistry that describes the concentration of a solution. It is defined as the number of moles of a solute present in one liter of solution. Molarity, often symbolized as \(M\), gives us insight into how densely packed the molecules are in the solution. This is especially useful when comparing concentrations across different solutions. To calculate molarity, use the formula \(M = \frac{\text{moles of solute}}{\text{liters of solution}}\). For example, if you have 1 mole of salt dissolved in 1 liter of water, the molarity would be 1 M. Understanding molarity allows you to prepare solutions with precise concentrations, which is vital for experiments and industrial applications.
Mixing Solutions
When you mix two or more solutions, it is important to consider the overall concentration and volume of the resulting mixture. The process often involves determining the total number of moles in the mixed solution and the total combined volume. In the given problem, we mixed two \(\mathrm{KMnO}_{4}\) solutions with different molarities and volumes. By calculating the moles of solute for each solution separately and then adding them, we determined the total moles of solute in the new mixture.It is crucial to remember that when mixing solutions, the volumes are additive, which means you simply add them together to find the total volume.
Mole Calculations in Solution
Mole calculations help to determine the number of molecules or atoms in a given quantity of a substance. In this context, we calculate the moles of the solute through the formula \(\text{moles} = \text{Molarity} \times \text{Volume (L)}\). This tells us how many moles of \(\mathrm{KMnO}_{4}\) are present in each solution individually before mixing. For example, in the first solution, with a molarity of 1.66 M and a volume of 0.0352 L, the moles are \(0.058432\). For the second, with 0.892 M and 0.0167 L, the moles are \(0.0148964\).These calculations are foundational when determining how much of a substance will react or be present in a mixture.
Volume Conversion Essentials
Volume conversion is an essential step when dealing with solutions in chemistry, particularly because solutions can be expressed in milliliters (mL) or liters (L). Often, calculations like those involving molarity require volumes in liters to ensure consistency with the SI unit system. To convert milliliters to liters, the conversion is straightforward: divide the volume in mL by 1000. For example, 35.2 mL becomes 0.0352 L. Doing this conversion allows you to correctly apply the molarity formula and achieve accurate results. Remember that correct volume conversion is a fundamental skill when mixing solutions or performing calculations to ensure your results are in the correct units.

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

Give the oxidation numbers for the underlined atoms in the following molecules and ions: (a) \(\mathrm{ClF},(\mathrm{b}) \mathrm{IF}_{7}\) (c) \(\underline{\mathrm{C}} \mathrm{H}_{4}\) (d) \(\underline{\mathrm{C}}_{2} \mathrm{H}_{2}\) (e) \(\underline{\mathrm{C}}_{2} \mathrm{H}_{4}\) (f) \(\mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4},(\mathrm{~g}) \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (h) \(\mathrm{KMnO}_{4}\), (i) \(\mathrm{NaHCO}_{3},(\mathrm{j}) \mathrm{Li}_{2},(\mathrm{k}) \mathrm{NaIO}_{3},\) (I) \(\mathrm{KO}_{2}\), \((\mathrm{m}) \mathrm{PF}_{6}^{-},(\mathrm{n}) \mathrm{K} \mathrm{AuCl}_{4}\)

Absorbance values for five standard solutions of a colored solute were determined at \(410 \mathrm{nm}\) with a \(1.00-\mathrm{cm}\) path length, giving the following table of data: \begin{tabular}{cc} Solute concentration \((M)\) & \multicolumn{1}{c} { A } \\ \hline 0.250 & 0.165 \\ 0.500 & 0.317 \\ 0.750 & 0.510 \\ 1.000 & 0.650 \\\ 1.250 & 0.837 \end{tabular} The absorbance of a solution of unknown concentration containing the same solute was 0.400 . (a) What is the concentration of the unknown solution? (b) Determine the absorbance values you would expect for solutions with the following concentrations: \(0.4 M, 0.6 M, 0.8 M\), 1\. \(1 M\). (c) Calculate the average molar absorptivity of the compound and determine the units of molar absorptivity.

What volume of \(0.112 M\) ammonium sulfate contains \(5.75 \mathrm{~g}\) of ammonium ion?

Arrange the following species in order of increasing oxidation number of the sulfur atom: (a) \(\mathrm{H}_{2} \mathrm{~S},(\mathrm{~b}) \mathrm{S}_{8}\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{S}^{2-}\) (e) HS \(^{-}\), (f) \(\mathrm{SO}_{2},(\mathrm{~g}) \mathrm{SO}_{3}\)

For each of the following pairs of combinations, indicate which one will produce the greater mass of solid product: a) \(105.5 \mathrm{~mL} 1.508 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(250.0 \mathrm{~mL}\) \(1.2075 \mathrm{M} \mathrm{KCl}\) or \(138.5 \mathrm{~mL} 1.469 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{~mL} 2.115 \mathrm{M} \mathrm{KCl}\) b) \(32.25 \mathrm{~mL} 0.9475 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(92.75 \mathrm{~mL} 0.7750 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) or \(52.50 \mathrm{~mL} 0.6810 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(39.50 \mathrm{~mL} 1.555 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) c) \(29.75 \mathrm{~mL} 1.575 \mathrm{M} \mathrm{AgNO}_{3}\) and \(25.00 \mathrm{~mL} 2.010 \mathrm{M}\) \(\mathrm{BaCl}_{2}\) or \(52.80 \mathrm{~mL} 2.010 \mathrm{M} \mathrm{AgNO}_{3}\) and \(73.50 \mathrm{~mL} 0.7500 \mathrm{M}\) \(\mathrm{BaCl}_{2}\)

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