In 91% (by volume) isopropyl alcohol (a large [compared to water], flammable molecule) solution, there are only two substances: the alcohol, and water. So when I poured 50 mL of 91% isopropyl alcohol into a graduated cylinder, how many milliliters were alcohol? How many were water? I’ll answer the first questions, putting a line through units that cancel. Then you answer the following questions doing the same.
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(91ml(iso) / 100ml(sol)) x 50ml(sol) = 45.5ml(iso) (because ml(sol) / ml(sol) =1, right?)
50 ml(sol) – 45.5ml(iso) = 4.5ml(water)
So, when I added 50 ml of water to the 91% isopropyl alcohol solution, how many total milliliters of water were in the solution?
To find the total milliliters of water in the solution after adding 50 ml of water to the 91% isopropyl alcohol solution, we need to calculate the new volume of water.
Given:
Initial volume of isopropyl alcohol = 45.5 ml
Volume of water added = 50 ml
To calculate the total volume of water in the solution, we can sum the initial volume of water in the solution and the volume of water added:
Total volume of water = initial volume of water + volume of water added
Total volume of water = 4.5 ml + 50 ml
Total volume of water = 54.5 ml
Therefore, there are 54.5 milliliters of water in the solution after adding 50 ml of water.
To find the total milliliters of water in the solution after adding 50 ml of water to the 91% isopropyl alcohol solution, you can use the formula:
Total milliliters of water = Initial milliliters of water + Milliliters of water added
Since the initial milliliters of water in the solution were 4.5 ml (as calculated in the previous step), and you added 50 ml of water, you can calculate:
Total milliliters of water = 4.5 ml + 50 ml = 54.5 ml
Therefore, after adding 50 ml of water to the 91% isopropyl alcohol solution, the total milliliters of water in the solution is 54.5 ml.