How much distilled water must be added to .85% solution Of NaCL to make a 70% solution of NaCl
You may have meant 85% but you typed .85%. You can't make 70% NaCl from 0.85% by adding water to it.
I meant 0.85% and 0.70% ...sorry
c1v1 = c2v2
To determine how much distilled water needs to be added, we need to consider the initial volume and concentration of the .85% solution of NaCl, as well as the desired concentration of the final 70% solution of NaCl.
Let's assume that you have an initial volume of 1 liter of the .85% NaCl solution. This means that the solution contains 0.85 grams of NaCl for every 100 milliliters of solution.
Now, let's calculate how much NaCl is present in the initial solution:
0.85% of 1000 milliliters (1 liter) = 0.85 grams
We want to end up with a 70% NaCl solution. This means that the final solution should contain 70 grams of NaCl for every 100 milliliters of solution.
Since we are adding distilled water, the amount of NaCl will remain the same (0.85 grams). We need to find out how much water we need to add to the initial solution.
Let's represent the amount of distilled water to be added as "W" (in milliliters).
The total volume of the final solution will be the sum of the initial volume (1 liter or 1000 milliliters) and the volume of the distilled water added (W milliliters).
To calculate the final concentration of the NaCl solution, we can use the following equation:
(0.85 grams) / (1000 + W milliliters) = (70 grams) / 100 milliliters
Now, we can solve this equation to find the value of W:
0.85 / (1000 + W) = 70 / 100
Cross-multiplying, we get:
(0.85) * 100 = (70) * (1000 + W)
85 = 70000 + 70W
Subtracting 70000 from both sides:
70W = -69815
Dividing by 70:
W ≈ -997
However, since we cannot add negative volume to the solution, it is not possible to make a 70% solution of NaCl using the given initial .85% NaCl solution.
Typically, when preparing a higher concentration solution, it is necessary to have a higher concentration initial solution, or to obtain a pure solid compound and dissolve it in the appropriate volume of solvent.