Determining the Mass of a Solution Containing 5% KI by Mass

Understanding the concentration of solutes in solutions is crucial in chemistry and various practical applications. This article will guide you through the process of calculating the mass of a solution containing 5% potassium iodide (KI) by mass, given a specific mass of KI. This is an important concept in chemistry, particularly in analytical chemistry and quantitative chemical analysis.

Problem Setup

The problem at hand involves a solution where the mass of potassium iodide (KI) is 258 mg. We need to determine the total mass of the solution that contains this amount of KI, with the KI concentration being 5% by mass. This translates to calculating the total mass of the solution, given that 5% of this mass is KI.

Step-by-Step Solution

1. **Understanding the Units**: First, it's crucial to understand the units used. 1000 mg is equal to 1 gram (g).

2. **Given Values**:

Mass of KI 258 mg 0.258 g Concentration of KI 5% by mass

3. **Setting Up the Equation**: We need to find the total mass of the solution (K g), where 5% of the solution mass is equal to the mass of KI. Mathematically, we express this as:

(frac{text{mass of KI}}{text{mass of solution}} times 100 5)

or simplified:

(frac{text{mass of KI}}{text{mass of solution}} frac{5}{100})

Substituting the given values into the equation:

(frac{0.258 , text{g}}{text{mass of solution}} frac{5}{100})

4. **Solving for the Mass of the Solution**: Rearrange the equation to solve for the mass of the solution:

(text{mass of solution} frac{0.258 , text{g}}{frac{5}{100}} 0.258 , text{g} times frac{100}{5} 5.16 , text{g})

Therefore, the total mass of the solution is 5.16 grams.

Conclusion

Understanding and being able to calculate the mass percent of a solute in a solution is critical for various applications, including chemical synthesis and pharmaceutical chemistry. The process involves using proportions and basic algebraic manipulation of mass calculations.

Related Concepts

These concepts are interconnected and form the foundation of more advanced chemical calculations. Familiarity with these concepts can aid in further studies of chemistry and related fields.

Additional Resources and Links

For further reading, explore the following resources:

Molarity and Concentration Calculations Solution Preparation