Fundamental Chemistry and Hydraulics - Mathematics and Statistics for the Environment - page 358

■ EXAMPLE 14.13
Problem: A chemical has a FW of 190 g/mol and we need 25 mL (0.025 L) of a 0.15- M solution.
How many grams of the chemical must be dissolved in 25 mL water to make this solution?

Solution:

Number grams/desired volume (L) = Desired molarity (mol/L) × FW (g/mol)

Rearranging,

Number grams = Desired volume (L) × desired molarity (mol/L) × FW (g/mol)

Number grams = 0.025 L × 0.15 mol/L × 190 g/mol = 0.7125 g/25 mL

14.1.2.8.5 Percent Solutions
Many reagents are mixed as percent solutions . When working with a dry chemical it is mixed as
dry mass (g) per volume, where the number of grams per 100 mL is the percent concentration.
A 10% solution is equal to 10 g dissolved in 100 mL of solvent. In addition, if we want to make
3% NaCl we would dissolve 3 g NaCl in 100 mL water (or the equivalent for whatever volume we
need). When using liquid reagents, the percent concentration is based upon volume per volume (e.g.,
number mL/100 mL). For example, if we want to make 70% ethanol we would mix 70 mL of 100%
ethanol with 30 mL water (or the equivalent for whatever volume we need). To convert from percent
solution to molarity, multiply the percent solution value by 10 to get grams per liter, then divide by
the formula weight:

(% Solution)
Formulaweight

×10

Molarity = (14.10)

■ EXAMPLE 14.14
Problem: Convert a 6.5% solution of a chemical with FW = 351 to molarity.

Solution:

( ) ×

6.5 g 100 mL
351 g/L

10

= 0 1852 . M

To convert from molarity to percent solution, multiply the molarity by the FW and divide by 10.

Molarity × Formulaweight

%Solution = (14.11)

10

■ EXAMPLE 14.15
Problem: Convert a 0.0045- M solution of a chemical having FW 176.5 to percent solution.

Solution:

0.0045 mol/L × 176.5 g/mol

= 00.%

10