How would you prepare 1.0 L of a 0.10 M solution of sulfuric acid from a 3.0 M solution of sulfuric acid?

1 Answer

This is an example of a dilution calculation. You can find two methods of doing dilution calculations here: How do you do dilution calculations?

In your problem,

#V_1# = ?; #c_1# = 3.0 mol/L
#V_2# = 1.0 L; #c_2# = 0.10 mol/L

Method 1. Using #V_1c_1 = V_2c_2#

#V_1c_1 = V_2c_2#

#V_1 = V_2 × c_2/c_1# = 1.0 L × #(0.10"mol/L")/(3.0"mol/L")# = 0.033 L = 33 mL

Method 2. Calculating moles

#V_2c_2# = 1.0 L × #(0.10"mol")/(1"L")# = 0.10 mol, so

#V_1c_1# = 0.10 mol

#V_1 = (0.10"mol")/(3.0"mol·L⁻¹")# = 0.033 L = 33 mL

In each case, you would place 33 mL of the concentrated solution in a 1 L volumetric flask. Then you would add enough distilled water to fill the flask to the 1 L mark