How many milliliters of glycerol, C3H8O3(d=1.26g/mL), must be added per kilogram of water to produce a solution with 4.85 mol % C3H8O3?
Kinda lost -.-;
Here is an answer I did last night.
http://www.jiskha.com/display.cgi?id=1266728125
the answer you did last night was wrong. sorry DrBOb :{
To solve this problem, we need to follow a step-by-step approach. Let's break it down:
Step 1: Convert mol % to grams
We are given that the solution should have 4.85 mol % C3H8O3. To calculate the grams of C3H8O3, we need to know the total weight of the solution.
Let's assume we have 1 kilogram (1000 grams) of water. The quantity of C3H8O3 in grams can be calculated as:
(4.85 mol / 100 mol) x 1000 g = 48.5 g
Step 2: Convert grams to milliliters
Now that we have the mass of C3H8O3, we can convert it to milliliters using its density. The density is given as 1.26 g/mL, which means that 1 mL of glycerol weighs 1.26 grams.
To find the volume in milliliters (mL), we divide the mass in grams (48.5 g) by the density (1.26 g/mL):
48.5 g / 1.26 g/mL = 38.49 mL
Therefore, approximately 38.49 milliliters of glycerol should be added per kilogram of water to produce a solution with 4.85 mol % C3H8O3.