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Calculate the volume of 0.123 -M \(\mathrm{NaOH}\) that contains \(25.0 \mathrm{~g} \mathrm{NaOH}\). Express your result in milliliters.

Short Answer

Expert verified
The volume required is 5080 mL.

Step by step solution

01

Determine the Molar Mass of NaOH

The molar mass of \(\mathrm{NaOH}\) can be calculated by adding the atomic masses of sodium \(\mathrm{(Na, 22.99\ g/mol)}\), oxygen \(\mathrm{(O, 16.00\ g/mol)}\), and hydrogen \(\mathrm{(H, 1.01\ g/mol)}\).\[\mathrm{Molar\ mass\ of\ NaOH} = 22.99\ g/mol + 16.00\ g/mol + 1.01\ g/mol = 40.00\ g/mol}\]
02

Calculate Moles of NaOH

Using the mass of \(\mathrm{NaOH}\), we calculate the number of moles using the formula: Moles = mass/molar mass. Given that the mass is \(25.0\ g\), the number of moles is\[\text{Moles of NaOH} = \frac{25.0\ g}{40.00\ g/mol} = 0.625\ mol\]
03

Use Molarity to Find Volume

We use the molarity formula to find the volume. Molarity (M) is defined as moles of solute per liter of solution: \(\text{Molarity} = \frac{\text{moles}}{\text{volume in liters}}\). Rearrange this to find volume: \(\text{Volume in liters} = \frac{\text{moles}}{\text{Molarity}}\). For a \(0.123\ M\) solution and \(0.625\ mol\):\[\text{Volume in liters} = \frac{0.625\ mol}{0.123\ M} = 5.08\ L\]
04

Convert Liters to Milliliters

Convert the volume from liters to milliliters by using the conversion factor \(1000\ mL = 1\ L\):\[\text{Volume in milliliters} = 5.08\ L \times 1000\ mL/L = 5080\ mL\]

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Key Concepts

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

Molarity
Molarity is a fundamental concept in chemistry that describes the concentration of a solution. It is defined as the number of moles of solute per liter of solution. Molarity is denoted by the symbol \(M\) and is used to convey how concentrated a solution is.

Understanding molarity helps chemists in preparing solutions with precision, essential for many chemical reactions. For a given solution, if you know the molarity and the volume, you can easily calculate the amount of solute present.

For instance, in our exercise, the molarity of the NaOH solution is given as 0.123 M. This indicates that there are 0.123 moles of NaOH in every liter of the solution. To find the volume of the solution needed to contain 0.625 moles of NaOH, we rearrange the molarity formula:
  • \( \text{Volume in liters} = \frac{\text{moles}}{\text{Molarity}} \)
With 0.625 moles and a molarity of 0.123 M, the calculated volume is 5.08 liters, which is then converted to milliliters.
Molar Mass
Molar mass is a key concept in chemistry used to convert between the mass of a substance and the amount in moles. It is defined as the mass of one mole of a given substance and is usually expressed in grams per mole (g/mol). Calculating the molar mass involves adding together the atomic masses of all the elements in a compound.

In our NaOH example, the molar mass is calculated as follows:
  • Sodium (Na): 22.99 g/mol
  • Oxygen (O): 16.00 g/mol
  • Hydrogen (H): 1.01 g/mol
The combined molar mass of NaOH is 40.00 g/mol. This means that one mole of sodium hydroxide weighs 40 grams.

In practice, knowing the molar mass allows us to determine the number of moles of a given mass of a substance. For example, using the molar mass of NaOH, we calculated that 25.0 grams of NaOH is equal to 0.625 moles.
Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It allows chemists to predict how much product will form based on the amount of reactants used.

In solution chemistry, stoichiometry often involves calculations using molarity, volume, and molar mass to find the amount of chemicals involved in a reaction. For example, our exercise is a straightforward application of stoichiometry: converting grams to moles and then using molarity to find the volume of solution needed.

Stoichiometry is crucial because it helps in:
  • Calculating reactant concentrations and product yield
  • Scaling reactions to desired product amounts
  • Balancing chemical equations to maintain the conservation of mass
In essence, stoichiometry connects the world of measurable mass with the molecular handling of chemicals, guiding us through precise and reproducible chemical procedures.

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Most popular questions from this chapter

An aqueous hydrochloric acid solution is \(37.0 \% \mathrm{HCl}\) by mass. The density of the solution is \(1.185 \mathrm{~g} / \mathrm{mL}\). Calculate the molarity of \(\mathrm{HCl}\) in this solution.

Determine which of these are redox reactions, which are acid-base reactions, and which are gas-forming reactions. Identify the oxidizing and reducing agents in each of the redox reactions. Identify the acid and base in each acidbase reaction. (a) \(\mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{NaH}_{2} \mathrm{PO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (b) \(\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{NH}_{4} \mathrm{HCO}_{3}(\mathrm{aq})\) (c) \(\mathrm{TiCl}_{4}(\mathrm{~g})+2 \mathrm{Mg}(\ell) \longrightarrow \mathrm{Ti}(\mathrm{s})+2 \mathrm{MgCl}_{2}(\ell)\) (d) \(\mathrm{NaCl}(\mathrm{s})+\mathrm{NaHSO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{HCl}(\mathrm{g})+\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\)

Determine the oxidation number of \(\mathrm{Mn}\) in each of thes species. (a) \(\left(\mathrm{MnF}_{6}\right)^{3-}\) (b) \(\mathrm{Mn}_{2} \mathrm{O}_{7}\) (c) \(\mathrm{MnO}_{4}^{-}\) (d) \(\mathrm{Mn}(\mathrm{CN})_{6}^{-}\) (e) \(\mathrm{MnO}_{2}\)

Predict whether each compound is soluble in water. Indicate which ions are present in solution for the watersoluble compounds. (a) Ammonium nitrate (b) Barium sulfate (c) Potassium acetate (d) Calcium carbonate (e) Sodium perchlorate

Calculate the volume of \(2.06-\mathrm{M} \mathrm{KMnO}_{4}\) that contains \(322 \mathrm{~g}\) solute.

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