Question

How many grams of KBr are contained in \(125 \mathrm{g}\) of a 6.25\% (by mass) KBr solution?

Short answer

The mass of KBr in the 125g solution with a 6.25% mass percentage can be found with the formula: \(mass\ of\ solute = (mass\ percentage\ ×\ mass\ of\ solution) / 100\). Plugging in the given values, we get \(mass\ of\ KBr = (6.25 × 125) / 100\), which simplifies to \(mass\ of\ KBr = 7.8125g\). Therefore, there are approximately \(7.81\mathrm{g}\) (rounded to two decimal places) of KBr in the given solution.

Step-by-step solution

Write down the given information

We have a KBr solution with a total mass (solution) of 125 g and a mass percentage of 6.25%.

Rearrange formula for mass percentage

We need to find the mass of KBr (solute). The formula for mass percentage is: mass percentage = (mass of solute / mass of solution) × 100 Rearranging for mass of solute, we get: mass of solute = (mass percentage × mass of solution) / 100

Plug in given values and solve

Now we can plug in the given values for mass percentage and total mass (solution) into the formula: mass of KBr = (6.25 × 125) / 100 mass of KBr = 781.25 / 100 mass of KBr = 7.8125g So, there are approximately \(7.81\mathrm{g}\) (rounded to two decimal places) of KBr in the given solution.

Key concepts

Molarity and Solution Concentration

Understanding molarity is essential in the world of chemistry as it relates to the concentration of a solution. Molarity, denoted as 'M,' is a unit of concentration that measures the number of moles of a solute that is present in one liter of solution. It tells us how much of a substance is dissolved in a specific volume of liquid, making it the standard unit for solution concentration in chemistry.

To calculate molarity, we divide the number of moles of the solute by the volume of the solution in liters. The formula for molarity can be expressed as:
\[ M = \frac{moles \, of \, solute}{liters \, of \, solution} \].

In practical applications, if you know the mass of your solute and the volume of your solution, you can first convert the mass into moles using the molar mass of the compound (found on the periodic table). Then, use the formula above to find the molarity. When dealing with real-world problems, it's crucial to pay attention to the units, as converting grams to moles and milliliters to liters is often required.

Stoichiometry

Stoichiometry is the backbone of chemical reactions, dealing with the quantitative relationships between the reactants and products in a chemical equation. It is a critical concept in chemistry that enables you to predict yields of products, determine reactant amounts needed for reactions, and calculate various components' relationships within a reaction.

One fundamental aspect of stoichiometry involves the mole ratio, derived from the balanced chemical equation. The mole ratio allows chemists to convert between moles of one substance to moles of another. To solidify this concept, consider a balanced equation such as:
\[2H_2 + O_2 \rightarrow 2H_2O\].

From this equation, we can see that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water. Stoichiometry also involves the conservation of mass: the mass of reactants used will equal the mass of products formed, following the law of conservation of mass. Additionally, stoichiometric calculations are a vital part of the chemical lab work, as they guide the measurement of reactants needed for experiments.

Chemistry Problem Solving

Chemistry problem solving involves a systematic approach to answering questions and drawing conclusions based on chemical properties and mathematical principles. To tackle a chemistry problem effectively, start by identifying the known and unknown variables. Then, determine which equations and concepts apply. It is essential to perform unit conversions where necessary and keep track of significant figures and rounding rules.

Always write down what you know, as in the case of the textbook problem, and apply the appropriate formula. Accurate problem solving in chemistry requires practice; so working through multiple problems will strengthen your understanding and familiarity with different types of calculations. When in doubt, consult the periodic table, remember to balance your equations, and don't hesitate to break down complex problems into simpler, more manageable parts. This approach will not just aid in solving textbook problems but also provides a solid foundation for real-world lab experiments and research.