Question

A chemist used 65.5 mL of a sodium phosphate stock solution to make up 0.675 M sodium phosphate in a 450 mL volumetric flask. What is the molarity of the original stock solution?

Answer 1

The concentration of the original stock solution is 0.09825M

Step-by-step explanation

Given that;

The initial volume of the stock solution is 450mL

The molarity of the solution prepared is 0.675M

The volume of the solution used is 65.5mL

Follow the steps below to find the molarity of the original stock solution.

Step 1; Use the dilution formula

`\text{ M1 V1 }=\text{ M2V2}`

Where

M1 is the initial molarity of the solution

V1 is the initial volume of the solution

M2 is the final concentration of the solution

V2 is the final volume of the solution.

Step 2; Convert the volume to liters from mL

Recall, that 1 mL is 0.001L

Hence,

V1 = 450 x 0.001

V1 = 0.45L

V2 = 65.5 x 0.001

V2 = 0.0655L

Step 3; Substitute the given data into the dilution formula in step 1

`\begin{gathered} \text{ M1 V1 }=\text{ M2V2} \\ 0.45\text{ }*\text{ M1}=\text{ 0.675}*0.0655 \\ \text{ 0.45M1 }=\text{ 0.0442} \\ \text{ Divide both sides by 0.45} \\ \frac{\cancel{0.45}* M1}{\cancel{0.45}}=\text{ }(0.0442)/(0.45) \\ \text{ M1 }=0.09825\text{ M} \end{gathered}`

Therefore, the concentration of the original stock solution is 0.09825M