00:01
Good day in this question we will be solving this using concentration so we are given masses here where mass of glucose c6 h12 06 given is 29 .2 grams mass of h2 o is equals to given is 355 grams so we solve for moles of which also so mole of c6 h12 .06 is equals to the mass divided by molar mass divided by 180.
00:48
.16 grams per mole this is equals to .162 mole next you have mole of h2o is equals to 355 grams divided by 18 .0 .0 .0.
01:07
This is equals to 19 .7 mole.
01:13
Next, we are asked to solve first for the molarity.
01:18
So, malarity is equals to mole of solute, which is mole of c6h12.
01:24
Divided by volume in liter of solution.
01:30
Substituting, solute here is 0 .162 mole, divided by the given volume of solution is 378 m.
01:39
On rating desolators where one liter is 1000 m l so molarity is equals to 0 .429m next solving for molality molarity is equals to mole of solute divided by kilogram of solvent so substituting mold of solute here is the leucose point 162 mole divided by the kilogram of solvent h2o 355 grams converting these to kilograms where one kilogram is 1 ,000 grams so molality is equals to 0 .456 small m next solving for the percent mass we have percent mass of glucose is equals to mass of glucose, 29 .2 grams, divided by the total mass, which is 29 .2 grams plus 355 grams, times 100%.
03:11
This is equals to 7 .60%.
03:24
And percent mass of h2o is equals to 355 grams divided by 29 .2 grams.
03:37
Plus 355 grams times 100%.
03:43
This is equals to 92 .40%...
