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 ?
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.
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?
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.
DrBob222 explained it well i think the first time. Unless 3.33ml is the wrong answer...
your all idiots!!! both .09505 and 3.3 are wrong were doing it in a computer program and we reworked the problem!!!
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.
To solve this problem, we need to first determine the mass of glycerol required per kilogram of water and then convert it to milliliters.
Step 1: Calculate the mass of glycerol required.
- The molecular weight of C3H8O3 (glycerol) is:
C: 3 × 12.01 g/mol = 36.03 g/mol
H: 8 × 1.01 g/mol = 8.08 g/mol
O: 3 × 16.00 g/mol = 48.00 g/mol
Total: 92.11 g/mol
- To find the mass of glycerol required per kilogram of water, we need to know the mole fraction of glycerol in the solution. The mole fraction is given as 4.55 mol%.
- Since mole fraction is the ratio of the moles of glycerol to the total moles of both glycerol and water, we can calculate the number of moles of glycerol as follows:
Moles of glycerol = (4.55 mol% / 100) × 1 kg = 0.0455 kg
- Now, we can find the mass of glycerol required using the molecular weight:
Mass of glycerol = moles of glycerol × molecular weight of glycerol
Mass of glycerol = 0.0455 kg × 92.11 g/mol = 4.169 g
Step 2: Convert the mass to volume in milliliters.
- We can use the density of glycerol, which is given as 1.26 g/ml, to convert the mass to volume.
- The volume of glycerol required is given by:
Volume of glycerol = Mass of glycerol / Density of glycerol
Volume of glycerol = 4.169 g / 1.26 g/ml = 3.303 ml
Therefore, you would need to add approximately 3.303 milliliters of glycerol per kilogram of water to produce a solution with 4.55 mol% C3H8O3.
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
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.