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How many milliliters of a \(0.350 \mathrm{M}\) KOH solution contain \(0.0171\) mol of KOH?

Short Answer

Expert verified
48.9 mL of the solution is needed.

Step by step solution

01

Understanding Molarity

Molarity is a measure of concentration representing the number of moles of solute per liter of solution. It is expressed as \( M = \frac{n}{V} \), where \( n \) is the number of moles and \( V \) is the volume in liters.
02

Identifying Given Values

We know the molarity \( M = 0.350 \) M, and the number of moles of KOH \( n = 0.0171 \) mol. We need to find the volume \( V \) in liters.
03

Applying the Molarity Formula

Rearrange the molarity formula to solve for volume: \( V = \frac{n}{M} \). Substitute the given values into this formula: \( V = \frac{0.0171}{0.350} \).
04

Calculating the Volume in Liters

Calculate the value: \( V = \frac{0.0171}{0.350} = 0.048857 \, ext{liters}\).
05

Converting Liters to Milliliters

Convert the volume from liters to milliliters by using the conversion \( 1 \, ext{liter} = 1000 \, ext{milliliters} \). Thus, \( 0.048857 \, ext{liters} \times 1000 = 48.857 \, ext{milliliters} \).
06

Rounding the Final Answer

Since significant figures are important, round the answer to three significant figures: 48.857 becomes 48.9 mL.

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Key Concepts

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

Concentration
Concentration is a way to describe how much solute is present in a solution compared to the total volume of the solution. It provides a quantitative measure of the strength or intensity of the solution. Molarity (M) is a common unit of concentration and is especially useful in chemistry. It is defined as the number of moles of solute divided by the volume of the solution in liters. For example, a molarity of 0.350 means that in every liter of solution, there are 0.350 moles of solute present.
  • High molarity indicates a concentrated solution with a lot of solute.
  • Low molarity indicates a dilute solution with less solute.
Having a clear understanding of concentration helps in predicting how a solution will behave in a chemical reaction or when it is mixed with other substances.
Moles of solute
Moles are a standard unit of measurement in chemistry that quantify the amount of substance. One mole is a specific number, known as Avogadro's number, approximately 6.022 x 10²³, which represents the number of atoms or molecules in one mole of a substance. Moles of solute are pivotal when it comes to calculations involving solutions. They allow you to identify precisely how many individual solute particles a solution contains.
  • Without moles, it would be challenging to accurately perform stoichiometric calculations for reactions.
  • Knowing the moles of solute can help set up the base for further calculations, like determining the required volume to achieve a specific concentration.
For example, in the exercise, the moles of KOH (0.0171 mol) tell us precisely how much solute we have, which can then be used in conjunction with the molarity to find the required volume of the solution.
Volume conversion
Volume conversion is the process of changing measurements from one volume unit to another, which is common in chemistry to ensure consistency and precision in results. Typically, volumes in chemistry are expressed in liters because of the relationship with molarity, but sometimes it is necessary to convert these to milliliters, especially when dealing with smaller quantities.
  • 1 liter is equal to 1000 milliliters, a conversion factor that conveniently transforms measurements.
  • Such conversions are important for laboratory settings where precise measurements are crucial.
In the exercise example, after calculating the volume of the solution in liters, it was converted to milliliters (from 0.048857 liters to 48.857 milliliters). Converting into milliliters made the answer more practical and easier to handle for amounts typically used in experiments or practical applications.

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Most popular questions from this chapter

Assign oxidation numbers to each element in the following ions: (a) \(\mathrm{ClO}_{3}\) (b) \(\mathrm{SO}_{3}^{2-}\) (c) \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) (d) \(\mathrm{NO}_{2}^{-}\) (e) \(\mathrm{BrO}^{-}\) (f) \(\mathrm{AsO}_{4}{ }^{3-}\)

Assume that you have an aqueous mixture of \(\mathrm{NaNO}_{3}\) and \(\mathrm{AgNO}_{3}\). How could you use a precipitation reaction to separate the two metal ions?

A sample weighing \(14.98 \mathrm{~g}\) and containing a small amount of copper was treated to give a solution containing aqueous \(\mathrm{Cu}^{2+}\) ions. Sodium iodide was then added to yield solid copper(I) iodide plus \(\mathrm{I}_{3}^{-}\) ion, and the \(\mathrm{I}_{3}^{-}\) was titrated with thiosulfate, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\). The titration required \(10.49 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) for complete reaction. What is the mass percent copper in the sample? The balanced equations are $$ \begin{aligned} &2 \mathrm{Cu}^{2+}(a q)+5 \Pi(a q) \longrightarrow 2 \operatorname{CuI}(s)+\mathrm{I}_{3}^{-}(a q) \\ &\mathrm{I}_{3}^{-}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}(a q) \longrightarrow 3 \mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q) \end{aligned} $$

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