Math Problem Statement

Determine the amount in grams of KCl that exists in 20.3 g of a solution that contains 1.14% KCl by mass.

Solution

In this problem, you are asked to determine the amount of potassium chloride (KCl) in grams within a solution. Here is the information given:

  1. The total mass of the solution is 20.320.3 grams.
  2. The solution contains 1.14%1.14\% KCl by mass.

Solution Steps

To find the mass of KCl in the solution, you can apply the concept of mass percentage:

Mass of KCl=(Percentage of KCl100)×Total mass of solution\text{Mass of KCl} = \left(\frac{\text{Percentage of KCl}}{100}\right) \times \text{Total mass of solution}

Plugging in the values:

Mass of KCl=(1.14100)×20.3\text{Mass of KCl} = \left(\frac{1.14}{100}\right) \times 20.3

Calculating this:

Mass of KCl=0.0114×20.3=0.23142grams\text{Mass of KCl} = 0.0114 \times 20.3 = 0.23142 \, \text{grams}

Rounding appropriately (based on the significant figures provided in the problem), the amount of KCl is approximately:

Mass of KCl0.231grams\text{Mass of KCl} \approx 0.231 \, \text{grams}

Answer

The solution contains approximately 0.231 grams of KCl.

Would you like further details on any part of this calculation?


Here are five additional questions to deepen your understanding:

  1. How would the result change if the solution contained 5% KCl by mass instead?
  2. How do you convert between mass percentage and concentration in mol/L for solutions?
  3. What is the molar mass of KCl, and how would you calculate moles of KCl from mass?
  4. How would you calculate the mass of KCl if you had 50 grams of this same solution?
  5. What are practical applications of determining mass percentages in chemistry?

Tip: Always keep an eye on significant figures; they reflect the precision of your measurements and calculations!

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Math Problem Analysis

Mathematical Concepts

Percentage
Dimensional Analysis

Formulas

Mass of solute = (Percentage by mass / 100) × Total mass of solution

Theorems

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Suitable Grade Level

High School (Grades 9-12)