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_1V1 = ?; c_1c1 = 3.0 mol/L
V_2V2 = 1.0 L; c_2c2 = 0.10 mol/L

Method 1. Using V_1c_1 = V_2c_2V1c1=V2c2

V_1c_1 = V_2c_2V1c1=V2c2

V_1 = V_2 × c_2/c_1V1=V2×c2c1 = 1.0 L × (0.10"mol/L")/(3.0"mol/L")0.10mol/L3.0mol/L = 0.033 L = 33 mL

Method 2. Calculating moles

V_2c_2V2c2 = 1.0 L × (0.10"mol")/(1"L")0.10mol1L = 0.10 mol, so

V_1c_1V1c1 = 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