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How many moles of \(\mathrm{MgCl}_{2}\) are present in \(60.0 \mathrm{~mL}\) of a \(0.100 \mathrm{M} \mathrm{MgCl}_{2}\) solution?

Short Answer

Expert verified
0.00600 moles of \(\mathrm{MgCl}_{2}\) are present.

Step by step solution

01

Understand Molarity

Molarity ( ext{M}) is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The formula to calculate moles when you have molarity is: \[ ext{{Moles}} = ext{{Molarity}} imes ext{{Volume in liters}} \]
02

Convert Milliliters to Liters

To use the molarity formula, the volume needs to be in liters. Convert the given volume from milliliters to liters:\[ 60.0 ext{{ mL}} = 0.0600 ext{{ L}} \]
03

Apply Molarity Formula

Using the formula from Step 1, calculate the number of moles of \(\mathrm{MgCl}_{2}\):\[ \text{{Moles of }} \mathrm{MgCl}_{2} = 0.100 ext{{ M}} \times 0.0600 ext{{ L}} \]
04

Calculate the Result

Multiply the molarity by the volume in liters to find the moles:\[ \text{{Moles of }} \mathrm{MgCl}_{2} = 0.100 \times 0.0600 = 0.00600 \text{{ moles}} \]
05

Verify the Units

Ensure that the units are correct. We started with molarity (moles per liter) and volume (liters), so the final unit of measure for the result should be moles, which it is. Thus, our calculations are consistent.

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Key Concepts

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

Concentration
Concentration is a key concept in chemistry that reflects the amount of a substance (solute) present in a given volume of solution. Think of concentration as the strength of a solution—it tells us how much solute is packed into a unit of solution. A common way to express concentration is in terms of molarity (M), defined as the number of moles of solute per liter of solution.
A more concentrated solution has more moles of solute per unit volume, while a less concentrated solution has fewer moles. This is critical in chemical reactions where the ratio of reactants affects the outcome.
  • Molarity (M) = Moles of solute / Volume of solution in liters
Concentration can also be expressed in different units such as percent concentration or parts per million (ppm), but molarity is often used in academic and many practical settings.
Understanding concentration helps in preparing solutions with precise chemical properties, which is crucial in laboratory experiments and industrial applications.
Conversion of units
Conversion of units is essential when dealing with chemical calculations, as it ensures the accuracy and consistency of results. Frequently, volumes are provided in milliliters (mL), but for molarity calculations, we need these volumes to be in liters (L). Converting units correctly is a fundamental skill in chemistry.
For example, to convert from milliliters to liters, you divide the number of milliliters by 1000, as there are 1000 milliliters in a liter. Here's the simple conversion formula:
  • Liters (L) = Milliliters (mL) / 1000
These unit conversions allow you to use formulas and calculations that require specific units, and ensure that your results are correct and meaningful. Always double-check your conversion to prevent errors in your calculations.
Chemical solutions
Chemical solutions are homogeneous mixtures where one substance (the solute) is dissolved in another (the solvent). In our exercise, \(\mathrm{MgCl}_{2}\) is the solute, and usually, water would be the solvent to create an aqueous solution. Understanding solutions and how they are prepared is critical in chemistry and numerous fields such as pharmacology and environmental science.
The properties of a solution depend on the type and amount of solute, the solvent used, and the temperature and pressure conditions of the system. Solutions have uniform composition and properties throughout, which is why they allow for consistent reactions and precise measurements.
  • Solute: The substance being dissolved (e.g., \(\mathrm{MgCl}_{2}\))
  • Solvent: The substance doing the dissolving (e.g., water)
  • Grasping the concept of chemical solutions enables you to predict how substances will behave when mixed, making it a cornerstone concept in various scientific applications.

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    Most popular questions from this chapter

    The concentration of lead ions \(\left(\mathrm{Pb}^{2+}\right)\) in a sample of polluted water that also contains nitrate ions \(\left(\mathrm{NO}_{3}^{-}\right)\) is determined by adding solid sodium sulfate \(\left(\mathrm{Na}_{2} \mathrm{SO}_{4}\right)\) to exactly \(500 \mathrm{~mL}\) of the water. (a) Write the molecular and net ionic equations for the reaction. (b) Calculate the molar concentration of \(\mathrm{Pb}^{2+}\) if \(0.00450 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) was needed for the complete precipitation of \(\mathrm{Pb}^{2+}\) ions as \(\mathrm{PbSO}_{4}\).

    Hydrogen halides \((\mathrm{HF}, \mathrm{HCl}, \mathrm{HBr}, \mathrm{HI})\) are highly reactive compounds that have many industrial and laboratory uses. (a) In the laboratory, HF and \(\underline{H C l}\) can be generated by combining \(\mathrm{CaF}_{2}\) and \(\mathrm{NaCl}\) with concentrated sulfuric acid. Write appropriate equations for the reactions. (Hint: These are not redox reactions.) (b) Why is it that \(\mathrm{HBr}\) and HI cannot be prepared similarly, that is, by combining NaBr and NaI with concentrated sulfuric acid? (Hint: \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a stronger oxidizing agent than both \(\mathrm{Br}_{2}\) and \(\mathrm{I}_{2} .\) ) \((\mathrm{c}) \mathrm{HBr}\) can be prepared by reacting phosphorus tribromide \(\left(\mathrm{PBr}_{3}\right)\) with water. Write an equation for this reaction.

    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 \(73.50 \mathrm{~mL} 0.7500 \mathrm{M}\) \(\mathrm{BaCl}_{2}\)

    Determine the resulting nitrate ion concentration when \(95.0 \mathrm{~mL}\) of \(0.992 \mathrm{M}\) potassium nitrate and \(155.5 \mathrm{~mL}\) of \(1.570 \mathrm{M}\) calcium nitrate are combined.

    Calculate the volume in milliliters of a solution required to provide the following: (a) \(2.14 \mathrm{~g}\) of sodium chloride from a \(0.270-M\) solution, (b) \(4.30 \mathrm{~g}\) of ethanol from a \(1.50-M\) solution, (c) \(0.85 \mathrm{~g}\) of acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) from a \(0.30-M\) solution.

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