Question

What volume of \(0.123 \mathrm{M} \mathrm{NaOH},\) in milliliters, contains \(25.0 \mathrm{g}\) of \(\mathrm{NaOH} ?\)

Short answer

5,081.3 mL of NaOH solution is needed.

Step-by-step solution

Calculate Moles of NaOH

First, we need to find out how many moles of NaOH are there in 25.0 grams of NaOH. The molecular weight of NaOH is approximately 40.0 g/mol. Use the formula:\[ \text{moles of NaOH} = \frac{\text{mass of NaOH}}{\text{molar mass of NaOH}} = \frac{25.0 \text{ g}}{40.0 \text{ g/mol}} = 0.625 \text{ moles} \]

Use Concentration to Find Volume

Now, use the concentration of the NaOH solution to find out the volume in liters. Concentration (Molarity) is defined as moles of solute per liter of solution. So, we rearrange the formula to find the volume:\[ \text{Volume (L)} = \frac{\text{moles of NaOH}}{\text{Molarity}} = \frac{0.625 \text{ moles}}{0.123 \text{ M}} \approx 5.0813 \text{ L} \]

Convert Volume to Milliliters

Finally, convert the volume from liters to milliliters by multiplying by 1000 (since 1 Liter = 1000 milliliters):\[ 5.0813 \text{ L} \times 1000 \text{ mL/L} = 5081.3 \text{ mL} \]

Key concepts

Molar Mass

Understanding molar mass is essential in stoichiometry. Molar mass, expressed in grams per mole (g/mol), quantifies the amount of mass in one mole of a substance. It is derived from the atomic masses found on the periodic table. For example, Sodium Hydroxide (NaOH) has a molar mass of approximately 40.0 g/mol, calculated by adding the atomic masses of sodium (Na), oxygen (O), and hydrogen (H).

• Sodium (Na) = 22.99 g/mol
• Oxygen (O) = 16.00 g/mol
• Hydrogen (H) = 1.01 g/mol

Therefore, \(\mathrm{NaOH}\) has a total molar mass of \( 22.99 + 16.00 + 1.01 = 40.00 \text{ g/mol} \). This molar mass is pivotal for converting the mass of a substance to moles, which is the next step in solving stoichiometry problems.

Molarity

Molarity (M) is a measure of concentration in chemistry, expressing the number of moles of a solute dissolved in one liter of solution. It is represented as \(\text{M} = \frac{\text{moles of solute}}{\text{liters of solution}}\). In the context of our problem, we are dealing with a NaOH solution with a molarity of 0.123 M.
To solve the problem, the molarity gives us a way to relate moles of solute to the volume of solution. Understanding molarity allows you to predict how solutions of a certain concentration will behave in reactions, determine reactant requirements in reactions, or calculate how much solvent is needed to dilute a concentrated solution. Here, it is used to find the volume of solution needed to provide a specific amount of NaOH.

Unit Conversion

Unit conversion is a fundamental skill in chemistry to ensure that all measurements are in their correct and most meaningful form. It involves converting one set of units to another. In our exercise, unit conversion is necessary to determine the volume of the NaOH solution in milliliters after initially calculating it in liters.
To convert between these units:

• Remember that 1 liter (L) is equal to 1000 milliliters (mL).
• Multiply the volume in liters by 1000 to obtain the result in milliliters.

So, a calculated volume of \(5.0813 \text{ L}\) converts to \(5.0813 \times 1000 = 5081.3 \text{ mL}\). Such conversions are critical since laboratory measurements often use milliliters rather than liters.

Mass-to-Mole Conversion

Converting mass to moles is a crucial process in stoichiometry. It allows chemists to translate the weight of a substance into moles, which is necessary for reacting substances accurately in the proper proportions. The conversion uses the formula: \[\text{moles} = \frac{\text{mass (in grams)}}{\text{molar mass (in g/mol)}}\].
In the given problem, you need to find the moles of NaOH from a known mass (25.0 grams). Given the molar mass of NaOH is 40.0 g/mol, the conversion can be performed as: \[\text{moles of NaOH} = \frac{25.0 \text{ g}}{40.0 \text{ g/mol}} = 0.625 \text{ moles}\]. Understanding this conversion process allows you to determine how much of a substance you have at the particle level, crucial for predicting and understanding chemical reactions.