Relative andabsolute salt concentration

[schillerteena]

I need desperate help in calculating the absolute and relative salt concentration. Does anyone know the formula to solve this? Please help

Make a stock solution of 1 cup of salt dissolved in 5 cups of water, as follows:
Pour 3 cups of water into your large container.
Add 1 cup of salt.
Stir to dissolve some of the salt. It will not all dissolve yet.
Add 2 more cups of water.
Stir to dissolve the rest of the salt. The salt should be completely dissolved before you go on to the next step.
This may take several (5 to 10) minutes of stirring, so you may need to be patient.
Make a two-fold serial dilution of the stock solution, as follows:
Label five of the plastic cups 1-5. Cup 1 will be for the stock solution, cups 2-4 will be for the dilutions, and cup 5 will be plain tap water.
Add 3/4 cup of your stock salt solution to cup 1.
Add 3/4 cup plain tap water to cups 2-5.
Measure out 3/4 cup stock solution, and add it to cup 2. Mix.
Measure out 3/4 cup of the solution from cup 2 and add it to cup 3. Mix.
Measure out 3/4 cup of the solution from cup 3 and add it to cup 4. Mix.
What are the relative salt concentrations of cups 1-4? Example: Cup 2 is made up of half stock solution and half tap water, which is a 50 percent relative salt concentration.
What are the absolute salt concentrations of cups 1-4? (If you want to convert to metric units, 1 cup of salt is about 292 grams [g], and 1 cup of water is 237 milliliters [mL].) Write these concentrations down in your lab notebook.

[rrv17]

Hello,

The equation to find concentration of a solution is as follows:

C=(m/v)

C is concentration, m is the mass of the solute in moles (the salt in this case), and v is the volume of the solution in liters (both the salt and the water combined together).

To get moles, divide the mass of the compound (the amount used as the solute), by the sum of the molecular mass of the compound. In this case, you would divide the grams of salt by the atomic mass of Na plus the atomic mass of Cl.

Hope you find this helpful, good luck!
-Regina V.