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GinnyAnswer
To prepare a 2.50 M HCl solution, you will need to dilute the 12.0 M HCl solution. Here's how you can calculate the volume of the 12.0 M HCl solution needed to prepare 1.25 L of a 2.50 M HCl solution: 1. Use the dilution formula: \(M_1V_1 = M_2V_2\), where: - \(M_1\) = initial concentration of the stock solution (12.0 M), - \(V_1\) = volume of the stock solution to be used (unknown), - \(M_2\) = final concentration of the diluted solution (2.50 M), - \(V_2\) = final volume of the diluted solution (1.25 L). 2. Plug in the values into the formula: \(12.0 M \times V_1 = 2.50 M \times 1.25 L\). 3. Solve for \(V_1\): \(12.0V_1 = 2.50 \times 1.25\), \(12.0V_1 = 3.125\), \(V_1 = \frac{3.125}{12.0}\), \(V_1 ā‰ˆ 0.26 L\). Therefore, you would need approximately 0.26 liters (260 mL) of the 12.0 M HCl solution to prepare 1.25 liters of a 2.50 M HCl solution through dilution.
SneezyDwarf125

Answer:

[tex]\boxed{\boxed{\huge\text{Volume of HCl$_{\,(aq)}$ needed $=0.260\ \rm L$}}}[/tex]

Explanation:

When preparing a solution by dilution, the total number of moles of solute remains constant before and after the dilution, We just add additional water (solvent) but the amount of solute does not change. Therefore since moles = concentration Ɨ volume, therefore we can use the dilution equation:

[tex]\boxed{\Large\text{$\rm c_1V_1=c_2V_2$}}[/tex]

[tex]\text{Where the subscripts 1 and 2 represent the values before and after dilution respectively.}[/tex]

Given:

  • Initial Concentration = 12.0 M
  • Final Concentration = 2.50 M
  • Final Volume = 1.25 L

Substituting our values into the dilution equation, and rearranging for Vā‚:

[tex]\Large\text{$\rm(12.0)V_1=(2.50)(1.25)$}\\[/tex]

[tex]\Large\text{$\rm\therefore V_1=0.260\ L$}[/tex]

Since the final volume was given in litres, we do not need to change the units to millilitres.

[tex]\boxed{\boxed{\Large\text{$\rm\therefore$ Volume of HCl solution needed $=0.260\ \rm L$}}}[/tex]

[tex]\hrulefill[/tex]

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