Question

The soda you are drinking contains \(0.5 \%\) by mass sodium benzoate as a preservative. What approximate mass of sodium benzoate is contained in \(1.00 \mathrm{~L}\) of the solution assuming that the density of the soda is \(1.00 \mathrm{~g} / \mathrm{mL}\) (the approximate density of water)?

Short answer

The approximate mass of sodium benzoate in 1.00 L of the soda solution is 5 g.

Step-by-step solution

Convert volume to milliliters

We are given 1.00 L of the solution. To convert this to milliliters, we use the conversion factor: \(1\mathrm{~L} = 1000\mathrm{~mL}\). So, we have: \(1.00\mathrm{~L} \times \dfrac{1000\mathrm{~mL}}{1\mathrm{~L}} = 1000\mathrm{~mL}\)

Determine the mass of the solution

The density of the soda is given as \(1.00 \mathrm{~g/mL}\). To find the mass of the solution, we use the formula: \(\text{mass} = \text{density} \times \text{volume}\) So, the mass of the solution is: \(\text{mass} = 1.00 \mathrm{~g/mL} \times 1000\mathrm{~mL} = 1000\mathrm{~g}\)

Calculate the mass of sodium benzoate

We are given that the mass percent of sodium benzoate in the soda is 0.5%. This means that for every 100 g of the solution, there are 0.5 g of sodium benzoate. We have 1000 g of the solution, so we can set up a proportion to find the mass of sodium benzoate, using the cross-multiplication technique: \(\dfrac{\text{mass of sodium benzoate}}{1000\mathrm{~g}} = \dfrac{0.5\mathrm{~g}}{100\mathrm{~g}}\) Now, solve for the mass of sodium benzoate: \(\text{mass of sodium benzoate} = \dfrac{0.5\mathrm{~g} \times 1000\mathrm{~g}}{100\mathrm{~g}} = 5\mathrm{~g}\) The approximate mass of sodium benzoate in 1.00 L of the solution is 5 g.

Key concepts

Sodium Benzoate

Sodium benzoate is a widely used food preservative that helps prevent the growth of mold, bacteria, and yeast in foods and beverages. It appears as a white, odorless powder. When dissolved, it inhibits the fermentation process, making it effective in food preservation, particularly in acidic environments such as sodas.

Its chemical formula is C₇H₅NaO₂, meaning each molecule contains seven carbon atoms, five hydrogen atoms, one sodium atom, and two oxygen atoms. This unique structure gives it preservative qualities.

Sodium benzoate is convenient because it is inexpensive and generally recognized as safe when used in limited quantities. However, it's crucial to adhere to recommended concentration levels in consumer products due to potential health concerns at high dosages.

Density of Solutions

Density is a measure of how much mass is contained in a given volume and is typically expressed in grams per milliliter (g/mL) or kilograms per liter (kg/L). It is an important property for liquids and solutions as it helps determine how substances mix and settle.

In this exercise, we assume the density of the soda is 1.00 g/mL. This is similar to water, making calculations straightforward. With this density, we expect that one milliliter of the solution weighs one gram.

Understanding density allows us to calculate the mass of a solution given its volume. Using the formula: \[ \text{mass} = \text{density} \times \text{volume} \] we simply multiply the given volume by the density to find the total mass of the solution in grams. This principle is crucial to finding components within a solution, such as sodium benzoate in beverages.

Conversion of Units

Conversion of units is fundamental in problem-solving, especially in chemistry, where various measurements must be translated into a coherent system. It involves changing a quantity expressed in one set of units to its equivalent in another set.

In the given exercise, we needed to convert liters to milliliters since density units were given in grams per milliliter (g/mL). The conversion factor used here is: \[ 1 \text{ L} = 1000 \text{ mL} \] This conversion makes the density and mass calculations possible by aligning the units correctly.

Whenever you perform unit conversions, remember:

• Write down the units at each step.
• Use a conversion factor that cancels the units you want to change.
• Check your final answer to ensure the units make sense in the context of the problem.

Mastering these conversions aids in precise scientific calculations and builds a solid foundation for more complex problems.