Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

How would you prepare \(500 \mathrm{~mL}\) of \(0.33 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) from solid \(\mathrm{CaCl}_{2} ?\) Specify the glassware that should be used.

Short Answer

Expert verified
Weigh 18.31 g of \( \mathrm{CaCl}_{2} \), dissolve in water, and dilute to 500 mL in a volumetric flask.

Step by step solution

01

Calculate Moles of Solute

To find the amount of calcium chloride needed, we first calculate moles using the formula \( n = M \times V \), where \( n \) is moles, \( M \) is molarity, and \( V \) is volume in liters. Here, \( M = 0.33 \; \mathrm{M} \) and \( V = 0.500 \; \mathrm{L} \). So, \( n = 0.33 \times 0.500 = 0.165 \; \mathrm{moles} \).
02

Convert Moles to Grams

Next, we convert moles of \( \mathrm{CaCl}_{2} \) to grams using the molar mass. The molar mass of \( \mathrm{CaCl}_{2} \) is approximately \( 110.98 \; \mathrm{g/mol} \). Thus, grams required = \( 0.165 \; \mathrm{moles} \times 110.98 \; \mathrm{g/mol} = 18.3127 \; \mathrm{g} \).
03

Weigh the Calcium Chloride

Use a balance to weigh \( 18.31 \; \mathrm{g} \) of \( \mathrm{CaCl}_{2} \) solid. Ensure the balance is calibrated before measuring for accuracy.
04

Dissolve Calcium Chloride in Water

Place the weighed \( \mathrm{CaCl}_{2} \) in a beaker and add distilled water gradually while stirring until the entire solid is dissolved.
05

Transfer and Dilute the Solution

Transfer the dissolved solution into a 500 mL volumetric flask. Rinse the beaker with distilled water adding the rinsings to the flask to ensure all solute is transferred. Add distilled water to the flask until the bottom of the meniscus is at the 500 mL mark.
06

Mix the Solution

Stopper the flask and mix by inverting the flask several times to ensure homogeneity of the 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.

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

Key Concepts

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

Calcium Chloride
Calcium chloride, with the chemical formula \( \mathrm{CaCl}_2 \), is a salt composed of calcium and chloride ions. It is highly soluble in water and widely used in different applications such as de-icing roads, as a drying agent, and in food preservation. In laboratories, calcium chloride is a common source of calcium ions and is used in the preparation of chemical solutions. When preparing a solution, it's crucial to measure an accurate amount of calcium chloride to ensure the desired concentration. This means starting with the correct number of moles, which are then converted to grams for practical use. Understanding the properties of calcium chloride, including its solubility and molarity in solutions, is key for effectively preparing accurate solutions.
Moles to Grams Conversion
Converting moles to grams is a fundamental concept in chemistry, crucial for preparing accurate chemical solutions. The conversion process involves using the molar mass of a substance, which is the mass of one mole of its molecules. For calcium chloride, the molar mass is approximately \( 110.98 \, \mathrm{g/mol} \).
  • First, calculate the number of moles needed using the equation \( n = M \times V \), where \( n \) is the number of moles, \( M \) is the molarity, and \( V \) is the volume in liters.
  • Once the moles are determined, convert them to grams by multiplying by the molar mass. This step translates theoretical calculations into a tangible amount of substance you can physically measure.
Converting moles to grams ensures that you have the exact amount of solute needed to achieve the desired concentration in your solution.
Volumetric Flask
A volumetric flask is an essential piece of glassware in preparing solutions with precise concentrations. It's designed with a flat bottom and a long neck with a single calibration mark that indicates a specific volume, such as 500 mL in this exercise.
  • The flask's precise calibration guarantees that the solution reaches exactly the intended volume, helping maintain the accuracy of the prepared solution's molarity.
  • To use it, after dissolving the solute in a smaller volume of water, transfer the solution into the volumetric flask.
  • Rinse any remaining traces from the initial container into the flask and add distilled water until the liquid hits the calibration mark. Always check the meniscus from eye level to ensure correct volume.
The volumetric flask ensures that your solution is consistent and reliable for experimental use.
Chemical Solution Concentration
Chemical solution concentration refers to the amount of solute present in a given volume of solvent, usually expressed in molarity (\( M \)). Molarity is defined as moles of solute per liter of solution, making it a key measurement in solution preparation.
  • Start by calculating the moles of solute required using \( n = M \times V \), where \( M \) is molarity and \( V \) is volume.
  • Convert these moles into a measurable quantity in grams as precise as possible. This step is crucial for maintaining the desired concentration.
  • Properly dilute the solution by ensuring the solute is entirely dissolved and accurately adjusted to the target volume in a volumetric flask. This guarantees a uniform concentration throughout the solution.
Understanding concentration helps in ensuring that the prepared solutions meet the experimental needs and produce reliable results.

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

If the following solutions are mixed, is the resulting solution acidic, basic, or neutral? (a) \(65.0 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{HClO}_{4}\) and \(40.0 \mathrm{~mL}\) of \(0.0750 \mathrm{M} \mathrm{NaOH}\) (b) \(125.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) and \(90.0 \mathrm{~mL}\) of \(0.0750 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\)

A \(1.268 \mathrm{~g}\) sample of a metal carbonate \(\left(\mathrm{MCO}_{3}\right)\) was treated with \(100.00 \mathrm{~mL}\) of \(0.1083 \mathrm{M}\) sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\), yielding \(\mathrm{CO}_{2}\) gas and an aqueous solution of the metal sulfate \(\left(\mathrm{MSO}_{4}\right)\). The solution was boiled to remove all the dissolved \(\mathrm{CO}_{2}\) and was then titrated with \(0.1241\) M NaOH. A \(71.02 \mathrm{~mL}\) volume of \(\mathrm{NaOH}\) was required to neutralize the excess \(\mathrm{H}_{2} \mathrm{SO}_{4}\). (a) What is the identity of the metal M? (b) How many liters of \(\mathrm{CO}_{2}\) gas were produced if the density of \(\mathrm{CO}_{2}\) is \(1.799 \mathrm{~g} / \mathrm{L} ?\)

Sodium nitrite, \(\mathrm{NaNO}_{2}\), is frequently added to processed meats as a preservative. The amount of nitrite ion in a sample can be determined by acidifying to form nitrous acid (HNO \(_{2}\) ), letting the nitrous acid react with an excess of iodide ion, and then titrating the \(\mathrm{I}_{3}^{-}\) ion that results with thiosulfate solution in the presence of a starch indicator. The unbalanced equations are (1) \(\mathrm{HNO}_{2}+\mathrm{I}^{-} \longrightarrow \mathrm{NO}+\mathrm{I}_{3}^{-}\) (in acidic solution) (2) \(\mathrm{I}_{3}^{-}+\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-} \longrightarrow \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}{ }^{2-}\) (a) Balance the two redox equations. (b) When a nitrite-containing sample with a mass of \(2.935 \mathrm{~g}\) was analyzed, \(18.77 \mathrm{~mL}\) of \(0.1500 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution was needed for the reaction. What is the mass percent of \(\mathrm{NO}_{2}^{-}\) ion in the sample?

What is the total molar concentration of ions in each of the following solutions? (a) A \(1.250 \mathrm{M}\) solution of \(\mathrm{CH}_{3} \mathrm{OH}\) (b) A \(0.225 \mathrm{M}\) solution of $\mathrm{HClO}_{4}

Ringer's solution, used in the treatment of burns and wounds, is prepared by dissolving \(4.30 \mathrm{~g}\) of \(\mathrm{NaCl}, 0.150 \mathrm{~g}\) of \(\mathrm{KCl}\), and \(0.165 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\) in water and diluting to a volume of \(500.0 \mathrm{~mL}\). What is the molarity of each of the component ions in the solution?

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.

Sign-up for free