Question

A solution containing \(12.0 \%\) sodium hydroxide by mass in water has a density of \(1.131 \mathrm{g} / \mathrm{mL}\). What volume of this solution, in liters, must be used in an application requiring \(2.25 \mathrm{kg}\) of sodium hydroxide?

Short answer

So, the volume of the sodium hydroxide solution required is approximately \(16.6 \, L\).

Step-by-step solution

Calculate Mass of Sodium Hydroxide in the Solution

From the exercise, the mass percentage of sodium hydroxide (NaOH) in the solution is \(12.0 \%\). This means that 100 g of the solution contains 12.0 g of NaOH. Thus, for a solution containing \(2.25 \, kg = 2250 \, g\) of NaOH, it can be calculated as: \[ \text{Mass of the solution} = \frac{\text{mass of NaOH}}{\text{% of NaOH by mass}} = \frac{2250 \, g}{0.12} = 18750 \, g\].

Calculate Volume of the Solution

\(\mathrm{Density} = \frac{\mathrm{Mass}}{\mathrm{Volume}}\), therefore we can solve for volume: \[ \mathrm{Volume} = \frac{\mathrm{Mass}}{\mathrm{Density}}=\frac{18750 \, g}{1.131 \, g/mL} = 16580.2 \, mL \].

Convert Volume to Liters

To convert the volume from milliliters (mL) to liters (L), remember that \(1 \, L = 1000 \, mL\). So, \[ \text{Volume} = 16580.2 \, mL × \frac{1 \, L}{1000 \, mL} = 16.58 \, L. \] The volume should be rounded to three significant figures since the least precise data given in the question was to three significant figures.

Key concepts

Mass Percentage

To start with understanding mass percentage, let's break it down. Mass percentage is a way of expressing the concentration of a component in a mixture. It indicates how many grams of a solute are present in 100 grams of a solution.

In the given exercise, the mass percentage of sodium hydroxide (NaOH) is 12.0%. This means that in every 100 grams of the total solution, there are 12 grams of NaOH. This provides a straightforward way to calculate how much solute is present if you know the total mass of your solution.

To find the total mass of the solution needed for 2.25 kg or 2250 grams of NaOH, you use the formula:

• Mass of the solution = Mass of NaOH / (Percentage of NaOH by mass)

Applying the numbers, it becomes:

\[ \text{Mass of the solution} = \frac{2250 \, g}{0.12} = 18750 \, g \]

Simply put, to find out how much solution you require, divide the mass of NaOH needed by the mass percentage in decimal form. This calculation reveals a total mass of 18750 grams required for the given application.

Density Calculation

Density is a key property in finding out how much space a certain mass occupies. It is defined as mass per unit volume and is usually expressed in grams per milliliter (g/mL) for liquids.

In our problem, the density of the sodium hydroxide solution is given as 1.131 g/mL. With density, we can determine the volume if we know the mass, using the formula:

• Volume = Mass / Density

Inserting the known values for our specific solution, the formula becomes:

\[ \text{Volume} = \frac{18750 \, g}{1.131 \, g/mL} = 16580.2 \, mL \]

This calculation helps convert mass to volume, ensuring we know how much physical space the solution will occupy. Thus, given the density, it is a pivotal step in identifying how to measure our solution.

Unit Conversion

Finally, understanding unit conversion is crucial in problem-solving, especially when dealing with different measurement systems. Converting units ensures that measurements are consistent and comparable.

In this specific exercise, the solution's volume is initially given in milliliters, but we need the final result in liters. This conversion is straightforward once you remember the basic relationship that 1 liter equals 1000 milliliters.

Use the formula:

• Volume in liters = Volume in milliliters × \( \frac{1 \, L}{1000 \, mL} \)

Plugging in our calculated volume:

\[ \text{Volume in liters} = 16580.2 \, mL \times \frac{1 \, L}{1000 \, mL} = 16.58 \, L \]

This ensures that we are working in the correct units, making the information useful and consistent with typical scientific practices. This step finalizes our calculation ensuring the solution provided meets the application's needs.