How many milliliters of 11.6 M hydrochloric acid solution, HCl (aq), are needed to obtain 0.255 moles of hydrochloric acid?
The correct answer and explanation is :
To calculate how many milliliters of an 11.6 M hydrochloric acid (HCl) solution are needed to obtain 0.255 moles of HCl, we can use the relationship between molarity, moles, and volume.
Key Concepts:
- Molarity (M): Molarity is the concentration of a solution, defined as the number of moles of solute per liter of solution. It is expressed in moles per liter (mol/L).
- Moles (mol): The amount of a substance in terms of moles.
- Volume (V): The amount of solution, typically measured in liters (L) or milliliters (mL).
The equation that relates molarity, volume, and moles is:
$$
\text{Molarity (M)} = \frac{\text{Moles of solute (mol)}}{\text{Volume of solution (L)}}
$$
We can rearrange this equation to solve for the volume of the solution:
$$
\text{Volume (L)} = \frac{\text{Moles of solute (mol)}}{\text{Molarity (M)}}
$$
Once we find the volume in liters, we can convert it to milliliters (mL) since 1 L = 1000 mL.
Step-by-Step Calculation:
- Given Values:
- Molarity of HCl solution = 11.6 M
- Moles of HCl needed = 0.255 mol
- Calculate the Volume in Liters:
$$
\text{Volume (L)} = \frac{0.255 \, \text{mol}}{11.6 \, \text{mol/L}} = 0.02196 \, \text{L}
$$
- Convert Volume to Milliliters:
Since 1 liter = 1000 milliliters:
$$
0.02196 \, \text{L} \times 1000 = 21.96 \, \text{mL}
$$
Final Answer:
To obtain 0.255 moles of hydrochloric acid, you would need 21.96 mL of an 11.6 M HCl solution.
Explanation:
- The molarity (M) of a solution tells us how many moles of solute are present in 1 liter of solution.
- By dividing the number of moles of solute you need (0.255 mol) by the molarity of the solution (11.6 M), you calculate the volume required to achieve that number of moles.
- Since the volume is typically given in liters, we multiply the result by 1000 to convert to milliliters, which is a more common unit for smaller volumes. This allows us to obtain 21.96 mL of solution to get the desired amount of HCl.