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What is the definition of molarity? Show that the ratio of millimoles (mmol) to milliliters (mL) is equivalent to the ratio of moles to liters.

Short Answer

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Molarity (M) is the number of moles of solute per liter of solution. The ratio of mmol to mL is equivalent to the ratio of moles to liters because each is a thousandth of the other.

Step by step solution

01

Define Molarity

Molarity (M) is defined as the number of moles of solute per liter of solution. It is expressed mathematically as: \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \).
02

Define Millimoles and Milliliters

One millimole (mmol) is one-thousandth of a mole, which means \( 1 \, mmol = 10^{-3} \, moles \). Similarly, one milliliter (mL) is one-thousandth of a liter, i.e., \( 1 \, mL = 10^{-3} \, liters \).
03

Show the Equivalence of Ratios

Using the conversion factors from the previous step, the ratio of millimoles to milliliters can be expressed as \( \frac{mmol}{mL} = \frac{10^{-3} \, moles}{10^{-3} \, liters} \). Simplifying the fraction by canceling out the \( 10^{-3} \) factor from numerator and denominator, we get \( \frac{moles}{liters} \), which is the molarity.

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Key Concepts

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

Understanding Moles of Solute
When studying chemistry, one of the most fundamental concepts you will encounter is the 'mole'. This is a unit of measurement used to express the amount of a chemical substance. In the context of solutions, the 'moles of solute' refers to the quantity of the substance that is dissolved to form the solution.

Remember that one mole is equivalent to Avogadro's number, which is approximately 6.022 x 10^23 particles, whether they're atoms, molecules, or ions. The importance of using moles is that it allows chemists to count individual particles through a quantity that can be measured in the lab.

Calculating the moles of solute will often be your first step in determining the concentration of a solution. Here is a simple way to calculate it:
  • Measure the mass of the solute in grams.
  • Determine the molecular weight of the solute in g/mol.
  • Divide the mass by the molecular weight to obtain the moles of solute.
Understanding this concept is essential for delving deeper into the study of molarity and solution concentrations.
Liters of Solution and Its Relation to Molarity
After understanding moles, the next step is to consider the volume of the solution where the solute is dissolved. This is typically measured in liters. 'Liters of solution' does not just refer to the amount of solvent used. It is the total volume of both solute and solvent after they have been combined to make the solution.

To measure this volume accurately, chemists often use volumetric flasks, which are calibrated to contain a precise amount of liquid when filled to the mark. In molarity calculations, accurate volume measurements are crucial because molarity is dependent on the total volume of solution, not just the solvent.

It's important to note that when preparing a solution, you should always add the solute to the solvent, not the other way around. Doing so ensures the solution's volume can be precisely controlled and measured. This detail is paramount in achieving the correct molarity for your solution.
Translating Millimoles to Milliliters for Practical Use
We often deal with smaller quantities in practical laboratory work. This is where the relationship between millimoles (mmol) and milliliters (mL) becomes pertinent. As you now know from the solution steps, a millimole is one-thousandth of a mole, and a milliliter is one-thousandth of a liter.

These smaller units make calculations more manageable when you work with small samples. By using millimoles and milliliters, you can easily apply the concepts of molarity to real-world situations without needing to convert large units.

The beauty of these units lies in their ratio. Since molarity can be described as moles per liter, the equivalent ratio in millimoles to milliliters remains consistent, thanks to the mathematics of scaling. This means that if you know the molarity of a solution, you can conveniently express it in millimoles per milliliter, if necessary, without changing the numeric value—only the units.

For example, a solution with a molarity of 1 M can also be described as having a concentration of 1 mmol/mL. This direct relationship simplifies the transition between lab-scale experiments and larger scale applications. Understanding this ratio is highly beneficial for practical experiments, dilutions, and scaling up reactions.

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Most popular questions from this chapter

Silver bromide is "insoluble." What does this mean about the concentrations of \(\mathrm{Ag}^{+}\) and \(\mathrm{Br}^{-}\) in a saturated solution of AgBr? What makes it possible for a precipitate of \(\mathrm{AgBr}\) to form when solutions of the soluble salts \(\mathrm{AgNO}_{3}\) and \(\mathrm{NaBr}\) are mixed?

A mixture was known to contain both \(\mathrm{KNO}_{3}\) and \(\mathrm{K}_{2} \mathrm{SO}_{3}\). To \(0.486 \mathrm{~g}\) of the mixture, dissolved in enough water to give \(50.00 \mathrm{~mL}\) of solution, was added \(50.00 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{HCl}\) (an excess of \(\mathrm{HCl}\) ). The reaction mixture was heated to drive off all of the \(\mathrm{SO}_{2}\), and then \(25.00 \mathrm{~mL}\) of the reaction mixture was titrated with \(0.100 \mathrm{MKOH}\) The titration required \(13.11 \mathrm{~mL}\) of the \(\mathrm{KOH}\) solution to reach an end point. What was the percentage by mass of \(\mathrm{K}_{2} \mathrm{SO}_{3}\) in the original mixture of \(\mathrm{KNO}_{3}\) and \(\mathrm{K}_{2} \mathrm{SO}_{3}\) ?

Write balanced ionic and net ionic equations for these reactions. (a) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q)+\mathrm{BaCl}_{2}(a q) \longrightarrow\) $$ \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{BaCO}_{3}(s) $$ (b) \(\mathrm{CuCl}_{2}(a q)+\mathrm{NaOH}(a q) \longrightarrow\) $$ \mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{NaCl}(a q) $$ (c) \(\mathrm{FeSO}_{4}(a q)+\mathrm{Na}_{3} \mathrm{PO}_{4}(a q) \longrightarrow\) \(\mathrm{Fe}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) (d) \(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{NiCl}_{2}(a q) \longrightarrow\) \(\mathrm{AgCl}(s)+\mathrm{Ni}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)\)

Butanoic acid (also called butyric acid), \(\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2},\) gives rancid butter its bad odor. What is the name of the salt \(\mathrm{NaC}_{4} \mathrm{H}_{7} \mathrm{O}_{2} ?\)

Write an equation for the ionization of hydrogen bromide, a molecular substance and a strong acid, in water.

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