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Deriving moles and volumes from molar concentrations

How much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from [link] ?

Solution

In this case, we can rearrange the definition of molarity to isolate the quantity sought, moles of sugar. We then substitute the value for molarity that we derived in [link] , 0.375 M :

M = mol solute L solution mol solute = M × L solution mol solute = 0.375 mol sugar L × ( 10 mL × 1 L 1000 mL ) = 0.004 mol sugar

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What volume (mL) of the sweetened tea described in [link] contains the same amount of sugar (mol) as 10 mL of the soft drink in this example?

Answer:

80 mL

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Calculating molar concentrations from the mass of solute

Distilled white vinegar ( [link] ) is a solution of acetic acid, CH 3 CO 2 H, in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?

A label on a container is shown. The label has a picture of a salad with the words “Distilled White Vinegar,” and, “Reduced with water to 5% acidity,” written above it.
Distilled white vinegar is a solution of acetic acid in water.

Solution

As in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, so we must use the solute’s molar mass to obtain the amount of solute in moles:

M = mol solute L solution = 25.2 g CH 3 CO 2 H × 1 mol CH 2 CO 2 H 60.052 g CH 2 CO 2 H 0.500 L solution = 0.839 M
M = mol solute L solution = 0.839 M M = 0.839 mol solute 1.00 L solution

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Calculate the molarity of 6.52 g of CoCl 2 (128.9 g/mol) dissolved in an aqueous solution with a total volume of 75.0 mL.

Answer:

0.674 M

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Determining the mass of solute in a given volume of solution

How many grams of NaCl are contained in 0.250 L of a 5.30- M solution?

Solution

The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in [link] :

M = mol solute L solution mol solute = M × L solution mol solute = 5.30 mol NaCl L × 0.250 L = 1.325 mol NaCl

Finally, this molar amount is used to derive the mass of NaCl:

1.325 mol NaCl × 58.44 g NaCl mol NaCl = 77.4 g NaCl

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How many grams of CaCl 2 (110.98 g/mol) are contained in 250.0 mL of a 0.200- M solution of calcium chloride?

Answer:

5.55 g CaCl 2

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When performing calculations stepwise, as in [link] , it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In [link] , the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g.

In addition to retaining a guard digit for intermediate calculations, we can also avoid rounding errors by performing computations in a single step (see [link] ). This eliminates intermediate steps so that only the final result is rounded.

Determining the volume of solution containing a given mass of solute

In [link] , we found the typical concentration of vinegar to be 0.839 M . What volume of vinegar contains 75.6 g of acetic acid?

Solution

First, use the molar mass to calculate moles of acetic acid from the given mass:

g solute × mol solute g solute = mol solute

Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:

mol solute × L solution mol solute = L solution

Combining these two steps into one yields:

g solute × mol solute g solute × L solution mol solute = L solution
75.6 g CH 3 CO 2 H ( mol CH 3 CO 2 H 60.05 g ) ( L solution 0.839 mol CH 3 CO 2 H ) = 1.50 L solution

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What volume of a 1.50- M KBr solution contains 66.0 g KBr?

Answer:

0.370 L

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Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
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