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?

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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?

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Answer 1 · 2014-07-09T00:14:50

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,

$V_{1}$ $V_1$ = ?; $c_{1}$ $c_1$ = 3.0 mol/L
$V_{2}$ $V_2$ = 1.0 L; $c_{2}$ $c_2$ = 0.10 mol/L

Method 1. Using $V_{1} c_{1} = V_{2} c_{2}$ $V_1c_1 = V_2c_2$

$V_{1} c_{1} = V_{2} c_{2}$ $V_1c_1 = V_2c_2$

$V_{1} = V_{2} \times \frac{c_{2}}{c_{1}}$ $V_1 = V_2 × c_2/c_1$ = 1.0 L × $\frac{0.10 mol/L}{3.0 mol/L}$ $(0.10"mol/L")/(3.0"mol/L")$ = 0.033 L = 33 mL

Method 2. Calculating moles

$V_{2} c_{2}$ $V_2c_2$ = 1.0 L × $\frac{0.10 mol}{1 L}$ $(0.10"mol")/(1"L")$ = 0.10 mol, so

$V_{1} c_{1}$ $V_1c_1$ = 0.10 mol

$V_{1} = \frac{0.10 mol}{3.0 mol\cdotL⁻¹}$ $V_1 = (0.10"mol")/(3.0"mol·L⁻¹")$ = 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

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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?

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