Question

Part II Dilution Problems: 17. 350.0 mL of water was added to a 2.3 L solution of NaCl. If the final concentration of the solution was 0.967 M, what was the original concentration of the solution?

Answer 1

The original concentration of the solution is 0.147 mol

Step-by-step explanation

Given that;

The volume of water is 350.0mL

The volume of NaCl solution is 2.3L

The final concentration of the solution is 0.967M

To find the original concentration of the solution, follow the steps below

Step1: Write the dilution formula

`\text{ M1V1}=M2V2`

Where

• M1 is the original concentration of the solution

,

• V1 is the original volume of the of the solution

,

• M2 is the final concentration of the solution

,

• V2 is the final volume of the solution

Step 2: Convert the volume of water to L

Recall, that 1mL is equivalent to 0.001L

`\begin{gathered} \text{ The volume can be converted below as} \\ \text{ 1mL }\rightarrow\text{ 0.001L} \\ \text{ 350mL }\rightarrow\text{ vL} \\ \text{ Cross multiply} \\ \text{ vL}*\text{ 1mL }=\text{ 350mL}*0.001L \\ \text{ Isolate v} \\ \text{ v }=\text{ }\frac{350\cancel{mL}*\text{ 0.001L}}{1\cancel{mL}} \\ \text{ v }=\text{ 350}*\text{ 0.001} \\ \text{ v }=\text{ 0.35L} \end{gathered}`

Step 3; Find the original concentration of the solution by substituting by the given data into the formula in step 1

`\begin{gathered} \text{ M1}*2.3\text{ }=\text{ 0.35}*\text{ 0.967} \\ \text{ 2.3M1 }=\text{ 0.33845} \\ \text{ Divide both sides by 2.3} \\ \text{ }(2.3M1)/(2.3)\text{ }=\text{ }(0.33845)/(2.3) \\ \text{ M1}=\text{ 0.147 mol} \end{gathered}`

Hence, the original concentration of the solution is 0.147 mol