Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 48

Given two liters of \(0.496 \mathrm{M} \mathrm{KCl},\) describe how you would use this solution to prepare \(250.0 \mathrm{mL}\) of \(0.175 \mathrm{M} \mathrm{KCl} .\) Give sufficient details so that another student could follow your instructions.

Short Answer

Expert verified
To prepare 250.0 mL of 0.175 M KCl from a 0.496 M KCl solution, this involves diluting 88.31 mL of the original solution with enough water to obtain a total volume of 250 mL.

Step by step solution

01

Setting up the Dilution Equation

Set up the dilution equation using the formula \(M1V1 = M2V2\), where M1 is initial molarity of the solution, V1 is the initial volume of the solution, M2 is final molarity of the solution, and V2 is final volume of the solution.
02

Substitution

Substitute the known values into the equation. The initial molarity (M1) is \(0.496 M\), the final molarity (M2) is \(0.175 M\), and the final volume V2 is \(250.0 mL\). The equation will be now changed to \(0.496 M \cdot V1 = 0.175 M \cdot 250.0 mL\).
03

Solving the Equation

Next, one needs to solve the equation for the unknown V1. Reframing the equation gives \(V1 = (0.175 M \cdot 250.0 mL) / 0.496 M\). As a result, we get that V1 is approximately \(88.31 mL\). This is the volume of \(0.496 M KCl\) solution required to prepare \(250 mL\) of \(0.175 M KCl\) solution.
04

Procedure Description to Perform the Dilution

To perform the dilution, the first step is to measure \(88.31 mL\) of the \(0.496 M KCl\) solution using a suitable measuring device like a graduated cylinder. Transfer the extracted \(0.496 M KCl\) solution to a container with a capacity of at least \(250 mL\). The remaining volume, which is about \(161.69 mL\) (calculated by subtracting \(88.31 mL\) from \(250 mL\)), should be filled up with distilled water. Stir the solution to ensure uniform distribution of the solute. With this procedure, we get \(250 mL\) of \(0.175 M KCl\) solution.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Solid calcium carbonate, \(\mathrm{CaCO}_{3}(\mathrm{s}),\) reacts with \(\mathrm{HCl}(\mathrm{aq})\) to form \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CaCl}_{2}(\mathrm{aq}),\) and \(\mathrm{CO}_{2}(\mathrm{g}) .\) If a \(45.0 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}(\mathrm{s})\) is added to \(1.25 \mathrm{L}\) of \(\mathrm{HCl}(\mathrm{aq})\) that is \(25.7 \% \mathrm{HCl}\) by mass \((d=1.13 \mathrm{g} / \mathrm{mL})\) what will be the molarity of \(\mathrm{HCl}\) in the solution after the reaction is completed? Assume that the solution volume remains constant.

A small piece of zinc is dissolved in \(50.00 \mathrm{mL}\) of \(1.035 \mathrm{M}\) HCl. At the conclusion of the reaction, the concentration of the \(50.00 \mathrm{mL}\) sample is redetermined and found to be \(0.812 \mathrm{M} \mathrm{HCl} .\) What must have been the mass of the piece of zinc that dissolved? $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$$

How many milliliters of \(0.650 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) are needed to precipitate all the silver in \(415 \mathrm{mL}\) of \(0.186 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) as \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s}) ?\) \(2 \mathrm{AgNO}_{3}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{CrO}_{4}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s})+2 \mathrm{KNO}_{3}(\mathrm{aq}) $$

What is the molarity of \(\mathrm{NaCl}(\mathrm{aq})\) if a solution has 1.52 ppm Na? Assume that NaCl is the only source of Na and that the solution density is \(1.00 \mathrm{g} / \mathrm{mL}\) (The unit \(p p m\) is parts per million; here it can be taken to mean g Na per million grams of solution.)

Excess \(\mathrm{NaHCO}_{3}\) is added to \(525 \mathrm{mL}\) of \(0.220 \mathrm{M}\) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} .\) These substances react as follows: \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaHCO}_{3}(\mathrm{s}) \longrightarrow\) $$ \mathrm{CuCO}_{3}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ (a) How many grams of the \(\mathrm{NaHCO}_{3}(\mathrm{s})\) will be consumed? (b) How many grams of \(\mathrm{CuCO}_{3}(\mathrm{s})\) will be produced?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation
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.

Sign-up for free