/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 109 What mass of \(\mathrm{KNO}_{3}\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What mass of \(\mathrm{KNO}_{3}\) is required to prepare \(125 \mathrm{g}\) of 1.5\% KNO \(_{3}\) solution?

Short Answer

Expert verified
To prepare a 1.5% KNO3 solution with a total mass of 125g, you would need 1.875g of KNO3.

Step by step solution

01

Write down the given information

We are given the total mass of the solution (125g) and the desired mass percentage of KNO3 (1.5%).
02

Use the formula for percentage mass concentration to set up the equation

Let the mass of KNO3 required be \(m\) (in grams). Using the percentage mass formula, we have: \(1.5\% = \frac{m}{125g} \times 100\)
03

Solve for the mass of KNO3 required

We need to solve the equation for \(m\): \(1.5 = \frac{m}{125} \times 100\) To isolate \(m\), let's divide both sides of the equation by 100: \(\frac{1.5}{100} = \frac{m}{125}\) Now, multiply both sides by 125: \(\frac{1.5}{100} \times 125 = m\) Evaluate the expression on the left side: \(m = 1.875\) So, the mass of KNO3 required is 1.875g.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Understanding stoichiometry is fundamental in mastering chemistry. The term stoichiometry is derived from the Greek words 'stoicheion' and 'metron,' which mean element and measure, respectively. At its core, stoichiometry deals with the quantitative relationships between the substances that are involved in chemical reactions.

In the context of solution preparation, stoichiometry allows us to calculate the exact amounts of reactants needed to form a desired product. By using stoichiometric calculations, one can determine the mass of a compound required to achieve a certain concentration in a given volume of solvent. These calculations are essential, particularly when preparing laboratory solutions or conducting any form of chemical synthesis.

For instance, if we are tasked with preparing a 1.5% mass percentage solution of KNO3, stoichiometry not only guides us in figuring out how much potassium nitrate (KNO3) we need but also ensures that the proportion of KNO3 in the solution aligns with the desired chemical reaction or application.
Solution Concentration
Solution concentration represents the amount of a substance (solute) present in a specific amount of solvent. The mass percentage concentration is a common way to express concentration, especially in chemistry. It is defined as the mass of the solute in a solution divided by the total mass of the solution (solute plus solvent) times 100%. This gives a percentage that shows the proportion of the solute in the mixture.

When we say that a solution has a 1.5% KNO3 concentration, we mean that for every 100 grams of the solution, 1.5 grams are KNO3. This is extremely useful when mixing solutions to ensure that the strength of the solution is appropriate for its intended use. Whether for a lab experiment, manufacturing process, or medication, getting the concentration right is crucial. To calculate the mass of KNO3 required for a 125g solution, we apply the concept of mass percentage which involves a straightforward multiplication and division, resulting in the determination of the solute's mass needed to create a solution of the specified concentration.
Chemical Calculations
Performing chemical calculations is integral to many areas within chemistry, from basic laboratory tasks to advanced research. Such calculations help us understand the proportions and quantities of chemicals involved in reactions and solution preparations.

To carry out these calculations effectively, one must be familiar with various concepts and formulas. In our exercise, we used the formula for calculating mass percentage concentration to determine the mass of KNO3 necessary to make a 125g solution with a 1.5% concentration.

These formulas are also underpinned by the conservation of mass principle, which states that matter is neither created nor destroyed in a chemical reaction. Hence, the mass we calculate for KNO3 includes all the atoms as they will exist in the final solution, ensuring accuracy and maintaining the balance of matter throughout the process.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When \(50 .\) mL of \(5.4 M \mathrm{NaCl}\) is diluted to a final volume of \(300 .\) mL, what is the concentration of \(\mathrm{NaCl}\) in the diluted solution?

Strictly speaking, the solvent is the component of a solution that is present in the largest amount on a mole basis. For solutions involving water, water is almost always the solvent because there tend to be many more water molecules present than molecules of any conceivable solute. To see why this is so, calculate the number of moles of water present in \(1.0 \mathrm{L}\) of water. Recall that the density of water is very nearly \(1.0 \mathrm{g} / \mathrm{mL}\) under most conditions.

If \(27.5 \mathrm{mL}\) of \(3.5 \times 10^{-2} \mathrm{N} \mathrm{Ca}(\mathrm{OH})_{2}\) solution is needed to neutralize \(10.0 \mathrm{mL}\) of nitric acid solution of unknown concentration, what is the normality of the nitric acid?

How many moles of the indicated solute does each of the following solutions contain? a. 1.5 L of \(3.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution b. 35 mL of 5.4 M NaCl solution c. 5.2 L of \(18 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution d. \(0.050 \mathrm{L}\) of \(1.1 \times 10^{-3} \mathrm{M}\) NaF solution

For each of the following, the mass of the solute is given, followed by the total volume of the solution prepared. Calculate the molarity. a. \(1.25 \mathrm{g} \mathrm{KNO}_{3} ; 115 \mathrm{mL}\) b. \(12.5 \mathrm{g} \mathrm{KNO}_{3} ; 1.15 \mathrm{L}\) c. \(1.25 \mathrm{mg} \mathrm{KNO}_{3} ; 1.15 \mathrm{mL}\) d. \(1.25 \mathrm{kg} \mathrm{KNO}_{3} ; 115 \mathrm{L}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.