If there were two solutions- one with areagent(100% pure) mixed in water and the other solution containing the same reagent (99.9% pure) mixed in equal volume of water. How significant would the difference in concentration be between these two solutions?
It all depends upon how accurate the results are to be. This is like asking a millionaire if the loss of a dime would be significant. Of course it would not. But if you asked a child for the only dime s/he has, do you think the chile will give it up or run away?
To determine the difference in concentration between the two solutions, we need to compare the amount of the reagent present in each solution.
For the first solution containing a 100% pure reagent mixed in water, it means that the entire solution is composed of the reagent. Therefore, the concentration of the reagent in this solution is 100%.
For the second solution containing a 99.9% pure reagent mixed in an equal volume of water, we know that 99.9% of the solution is composed of the reagent. Considering the equal volume of water, it means that the concentration of the reagent in this solution is 49.95%.
To calculate this, we divide the concentration of the reagent by the total volume of the solution (reagent + water) and multiply by 100:
Concentration (in %) = (concentration of reagent / total volume of solution) * 100
For the second solution with a 99.9% pure reagent, 99.9% is the concentration of the reagent, and since it is mixed in an equal volume, the total volume of the solution is two times the volume of reagent. So, the concentration in % would be:
Concentration (in %) = (99.9 / 2) * 100 = 49.95%
Therefore, the difference in concentration between these two solutions would be approximately 50% (100% - 49.95% = 0.05%).