{"id":6895,"date":"2014-04-09T20:38:39","date_gmt":"2014-04-09T20:38:39","guid":{"rendered":"https:\/\/schooltutoring.com\/help\/?p=6895"},"modified":"2014-12-02T08:25:30","modified_gmt":"2014-12-02T08:25:30","slug":"how-to-prepare-a-solution-of-proper-concentration","status":"publish","type":"post","link":"https:\/\/schooltutoring.com\/help\/how-to-prepare-a-solution-of-proper-concentration\/","title":{"rendered":"How to Prepare a Solution of Proper Concentration"},"content":{"rendered":"

Overview:<\/b><\/h3>\n

When in a chemistry lab or just writing a test, many students struggle with how to create a solution of known concentration. Students find it difficult because they try to calculate the solution in terms of moles, which is a unit that cannot be measured in the lab. There are some easy to follow steps on how to prepare a proper solution with a known concentration.<\/p>\n

Step 1: Determining the number of moles of compound<\/b><\/h3>\n

The first step is to determine what concentration of solution you need to make along with the volume you are looking to end up with. In this example we are going to make 1234 mL of a 1.54 molar solution of NaCl (Sodium chloride).<\/p>\n

First, we will determine the number of mols of NaCl that are contained in 1234mL of a 1.54 molar solution. To do this, we will use the following formula:<\/p>\n

\"Equation1\"<\/a><\/p>\n

Where:<\/p>\n

n = number of moles of solution in mol<\/p>\n

C = concentration of solution in mol\/L<\/p>\n

V = volume of solution in L<\/p>\n

NOTE:<\/b><\/span> Always ensure that your units are consistent.<\/p>\n

Note that the unit of volume is mL while the unit for C is mol\/L. This means that we need to convert mL to L before solving the equation. 1234 mL is the same at 1.234 L. Now that all of our units are consistent, we can substitute our known values for concentration and volume:<\/p>\n

Step 2: Determining the mass of the compound<\/b><\/h3>\n

We now know that we need to add 1.90 mol of NaCl to our solution; however, mol is not a unit we can measure out with common lab instruments. We must first convert it to a unit that is easily measured in the lab. This requires a new equation:<\/p>\n

\u00a0\"Equation2\"<\/a><\/p>\n

Where:<\/span><\/p>\n

n = number of mole in mol<\/p>\n

m = grams of compound in g<\/p>\n

M = molar mass of compound in g\/ mol ( This is taken directly from the periodic table)<\/p>\n

To determine M, we need to take a closer look at our compound:<\/p>\n

NaCl \u00a0 \u00a0 \u00a0 \u00a0 = 1 Na + 1 Cl<\/p>\n

\u00a0= (1 * 22.99 g\/mol) + (1 * 35.45 g\/mol)<\/p>\n

\u00a0= 58.44 g\/ mol<\/p>\n

We have determined our molar mass (M) and calculated n in Step 1. We are now ready to use substitution to find m. Since all of our units are consistent, no further adjustments need to be made.<\/p>\n

m \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = n * M<\/p>\n

\"Equation3\"<\/a><\/p>\n

We now know that we must measure out 111.0 grams of NaCl on a scale for it to be added to solution.<\/p>\n

NOTE:<\/b> <\/span>The key when creating any solution is to slowly add the compound to water while stirring. This will start off with a very low concentration of the compound and slowly increase to the desired concentration. \u00a0If water is added to the compound instead, this will start off with a very high concentration of the solution, potentially resulting in a hazardous compound.<\/p>\n

Fill a beaker about \u00be full with deionized water, then add the 111.0 g of NaCl. After the NaCl is added, fill the container the rest of the way to the 1.234L mark. You now have 1.234 L of a 1.54 molar solution of NaCl.<\/p>\n

Interested in\u00a0chemistry tutoring services<\/a>? Learn more about how we are assisting thousands of students each academic year.<\/p>\n

SchoolTutoring Academy<\/span><\/strong>\u00a0is the premier educational services company for K-12 and college students. We offer tutoring programs for students in K-12, AP classes, and college. To learn more about how we help parents and students in Norfolk, VA: visit\u00a0Tutoring in Norfolk, VA<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Overview: When in a chemistry lab or just writing a test, many students struggle with how to create a solution of known concentration. Students find it difficult because they try to calculate the solution in terms of moles, which is a unit that cannot be measured in the lab. There are some easy to follow […]<\/p>\n","protected":false},"author":17,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"inline_featured_image":false,"footnotes":""},"categories":[5,16,1],"tags":[],"class_list":["post-6895","post","type-post","status-publish","format-standard","hentry","category-chemistry","category-science","category-uncategorized"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts\/6895","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/users\/17"}],"replies":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/comments?post=6895"}],"version-history":[{"count":0,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts\/6895\/revisions"}],"wp:attachment":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/media?parent=6895"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/categories?post=6895"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/tags?post=6895"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}