Concentration expressed in terms of molarity

1.  You want to make 2 liter solution with a concentration of 1,4 mol/dm3 KH2PO4. How much KH2PO4 do you need for this?

To solve this problem we have to answer the following questions:

-  What does it mean that a solution concentration is 1,4 mol/dm3 ?

-  What is the molar mass of KH2PO4 ?

-  How many moles of KH2PO4 are in 2 liter solution?

-  You many gramms of KH2PO4 are in 2 liter solution?

A solution concentration is 1,4 mol/dm3 means:

In 1dm3 solution there is 1,4 mol of solute.

If 1 dm3 solution contains 1.4 mol KH2PO4

Then 2 dm3 solution contains X mol KH2PO4

=2.8mol

MKH2PO4 = 136g/mol

1 mol KH2PO4 136g

2.8 mol KH2PO4 x g

x = 380.08g

o  2. Make 250,0 cm3 0,2 molar (mol/dm3) KNO3 solution!

Result: 5,05 g KNO3 is needed

o  3. Make 500,0 cm3, 0,2 molar (mol/dm3) KMnO4 solution

.

Result: 15,8 g KMnO4 is needed

4.

250 cm3 solution contains 2 g NaOH, what is the molarity of this solution?

250 cm3 solution contains 2 g NaOH

1000 cm3 solution contains x g NaOH

x = 8 g

1 mol NaOH 40 g

x mol NaOH 8 g

x = 0,2 mol

1000 cm3 solution contains 0,2 mol NaOH, therefore c = 0,2 mol/dm3 .

o  5. What is the molarity of 500,0 cm3solution containing 2,00 g NaNO3,?

Result: c = 0,05 mol/dm3

o  6.What is the molarity of 200,0 cm3solution containing 11,7 g NaCl?

o  Result: c = 1 mol/dm3

Concentration expressed in terms of percentage

o  Source: http://www.physiologyweb.com/calculators/percent_solutions_calculator.html

o  It is very common to express the concentration of solutions in terms of percentages. Percent means per 100 parts, where for solutions, part refers to a measure of mass (μg, mg, g, kg, etc.) or volume (μL, mL, L, etc.). In percent solutions, the amount (weight or volume) of a solute is expressed as a percentage of the total solution weight or volume. Percent solutions can take the form ofweight/volume %(wt/vol % or w/v %),weight/weight %(wt/wt % or w/w %), orvolume/volume %(vol/vol % or v/v %). In each case, the percentage concentration is calculated as the fraction of the weight or volume of the solute related to the total weight or volume of the solution.

o  Because percent solutions can be expressed in three different ways, it is imperative that the type of percent solution be explicitly stated. If this information is not provided, the end user is left to "guess" whether w/v %, w/w %, or v/v % was used. Each percent solution is appropriate for a number of different applications. For example, commercial aqueous reagents, such as concentrated acids and bases, are typically expressed as weight/weight % solutions. For example, commercially available concentrated hydrochloric acid (HCl) is 37% by weight (w/w %). On the other hand, many dilute solutions used for biological research are expressed as weight/volume % (e.g., 1% sodium dodecyl sulfate, SDS). Volume/volume % solutes are also common, and are used when pure solutes in liquid form are used. For example, a 70 % (v/v) solution of ethanol can be prepared by dissolving 70 mL of 100% (i.e., 200 proof) ethanol in a total solution volume of 100 mL.

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o  7.

o  Make 3,5 dm3 KNO3 with concentration of 20 w/V% .

100 cm3 solution contains 20g KNO3

3500 cm3 solution contains x g KNO3

x = 700 g

8.

700g solution contains 34g NaCl . What is its concentration in w/w%?

700g contains 34g NaCl

100g contains x g KNO3

x = 4.86 g

Thus the solution’s concentration is 4.86 w/w%.