2.1. Whatvolume (in ml) of a sulfuric acid solution with a molar concentration of 17.47mol/Ishould be taken to prepare1 liter of a solution with a molar concentration of sulfuricacid equivalent approximately equal to 0.1mol/I The answershould be presented with accuracy: X,X

To solve this problem, we need to use the concept of dilution and the formula for dilution:<br /><br />\[ C_1V_1 = C_2V_2 \]<br /><br />where:<br />- \( C_1 \) is the initial concentration of the solution,<br />- \( V_1 \) is the volume of the initial solution needed,<br />- \( C_2 \) is the final concentration of the solution,<br />- \( V_2 \) is the final volume of the solution.<br /><br />Given:<br />- \( C_1 = 17.47 \, \text{mol/L} \)<br />- \( C_2 = 0.1 \, \text{mol/L} \)<br />- \( V_2 = 1 \, \text{L} \)<br /><br />We need to find \( V_1 \).<br /><br />Using the dilution formula:<br /><br />\[ V_1 = \frac{C_2 \times V_2}{C_1} \]<br /><br />Substitute the given values:<br /><br />\[ V_1 = \frac{0.1 \, \text{mol/L} \times 1 \, \text{L}}{17.47 \, \text{mol/L}} \]<br /><br />\[ V_1 = \frac{0.1}{17.47} \]<br /><br />\[ V_1 \approx 0.0057 \, \text{L} \]<br /><br />Convert \( V_1 \) to milliliters (1 L = 1000 mL):<br /><br />\[ V_1 \approx 0.0057 \, \text{L} \times 1000 \, \text{mL/L} \]<br /><br />\[ V_1 \approx 5.7 \, \text{mL} \]<br /><br />So, the volume of the sulfuric acid solution with a molar concentration of \( 17.47 \, \text{mol/L} \) that should be taken to prepare 1 liter of a solution with a molar concentration of sulfuric acid equivalent approximately equal to \( 0.1 \, \text{mol/L} \) is approximately \( 5.7 \, \text{mL} \).