given that 5milimeter of 5.2mol per liter NaCl will be used to prepare 0.026mol per liter NaCl, describe how you will prepare this solution?
you are diluting it 5.2/.026=200 times, so you need one part orignal NaCl, 199 parts water. If one part is 5mL, then you add water to the 200*5ml=1000ml mark.
thanks so much
To prepare a solution with a concentration of 0.026 mol per liter NaCl using a 5.2 mol per liter stock solution, you can follow the steps below:
1. Determine the volume of the stock solution needed:
The volume can be calculated using the equation:
Volume (V1) x Concentration (C1) = Volume (V2) x Concentration (C2)
where V1 is the volume of the stock solution, C1 is the concentration of the stock solution, V2 is the volume of the final solution, and C2 is the desired concentration.
Rearranging the equation to solve for V1:
V1 = (V2 x C2) / C1
Plugging in the given values:
V1 = (0.026 mol/L x 5 mL) / 5.2 mol/L
V1 ≈ 0.025 L (or 25 mL)
Therefore, you will need 25 mL of the 5.2 mol per liter stock solution.
2. Measure and transfer 25 mL of the 5.2 mol per liter NaCl solution using a graduated cylinder or pipette.
3. Add the measured 25 mL of the stock solution to a container, such as a beaker or volumetric flask.
4. Add sufficient water to the container to bring the total volume up to 1 liter, ensuring that the final volume is accurately measured.
5. Stir or mix the contents of the container thoroughly to ensure uniform distribution of the solute.
6. Label the container with the concentration, date prepared, and any other relevant information.
By following these steps, you will have successfully prepared a solution with a concentration of 0.026 mol per liter NaCl using the given stock solution.