How many milliliters of glycerol, d=1.26 g/ml, must be added per kilogram of water to produce a solution with 4.55 mol% C3H8O3 ?

Answer 1

4.55 mole percent = 0.0455 mole fraction.
So you want 0.0455 mole glycerin/kg water.
Do you know what mole fraction is?
mole fraction glycerin = moles gly/total moles where total moles = moles glycerin + moles water.
moles water in 1 kg = 1000/molar mass
moles glycerin = 0.0455

mole fraction glycerin = 0.0455/(0.0455 + 55.5) = ??
Convert 0.0455 moles glycerin to grams and convert grams to mL using the density.

Answer 2

Im still not sure what to do. If i convert the .0455 moles into grams and then use the density to find ml, how does that take into account my 1kg of water?

Answer 3

The 1 kg of water is 1000 g/18.015 = 55.5093 moles water. Add that to 0.0455 moles solute to make total moles = 55.5548. So mole fraction 0.0455/55.5548 = 8.19 x 10^-4 glycerin. Now convert moles glycerin to grams and from there to mL using density.

Answer 4

What the hec are you tlaking about the answer does not come out, can you please solve it out and post an answer so we can check our work.

Answer 5

DrBob222 explained it well i think the first time. Unless 3.33ml is the wrong answer...

Answer 6

your all idiots!!! both .09505 and 3.3 are wrong were doing it in a computer program and we reworked the problem!!!

Answer 7

4.85 mole % means 4.8 mole of glycerol per 100 mol of water
To make 4.85 mol% of glycerol in 1800g of water, we need 4.85x 92g = 446.2 g of glycerol
So in 1000g (1kg) water we need 247.8 g of glycerol ( 446.2x1000/1800)
Density = mass/volume
Volume = mass/density
= 247.8/1.26
=196.66 mL

Answer 8

Renjith is right, except that we do not have 100 mol of H20. We assume 100 mol for the total of our solution but the mol of H20 is 100 - 4.85 mol glycerol = 95.15 mol glycerol. Do your calculations with this number instead, and you'll be fine.