A 4510 g sample of ethylene glycol c2h6o2 is dissolved in...

A 45.10 g sample of ethylene glycol (C2H6O2) is dissolved in 250.0 g of water at 25.0 °C. Determine the molarity (M), mole fraction of ethylene glycol (χ), molality (m), percent mass to volume (w/v)%, percent mass to mass (w/w)%, and percent volume to volume (v/v)%. Molar mass of ethylene glycol = 62.10 g/mol; density of EG = 1.11 g/mL; density of solution = 1.10 g/mL. Show all work using dimensional analysis and report your final answers in the spaces provided. Be mindful of SFs and rounding errors.

Transcript

00:01 Hello students for this question part a we have to calculate the molarity okay so for this the number of moles of solute present in thousand eml solution is called the molarity so here we have given 45 .10 gram of ethylene glycol is equal to 45 .10 .60 .62 .10 mole okay so the total mass total mass will be equal to 295 .10 gram so it is 295 .10 gram.

00:49 How we get it? we get it 45 .10 we add with 250 gram.

00:57 So you get this one.

00:58 Okay then the density of the solution density of the solution is given 1 .5.

01:04 1 .10 gram per emm.

01:07 So now 245 .10 solution will be 295 .10 upon 1 .10 ml will be equal to 268 .27m.

01:29 Okay.

01:30 Then you can write 268 .27ml.

01:31 Okay.

01:31 Then you can write 268 .27ml.

01:37 Solution contain 45 .10 upon 62 .10 that is equal to 45 .10.

01:48 At more of glycol.

01:55 Okay.

01:56 Then after that, 268 solution containing 268 .27 solution containing, okay, i'm explaining above.

02:10 So after this you can write that thousand ml of solution containing 45 .10 that multiplied by thousand upon 62 .10 multiplied by 268 .27 so you get answer 2 .7071 mole or 2 .71 mode you also can write it okay then after that the molarity of solution will be 2 .71 mold okay so this is for the part a okay so after this i will explain the part b okay so i need to clear this one wait then after i am explaining the three that is a molality okay so the number of mole of cellules 100 gram solvent is called the molality so 250 gram of solution contain 45 .10 upon 62 .10 mole of ethylene and then thousand gram of the solution contain 45 .10 into 4 upon 62 .10 that is 2 .90 mole then it is molality will be molality will be 2 .90 mole this is the morality then again i'm explaining the part two that is called the mole fraction it is asked so for the mole fraction what we do that we know that the number of mole of substances divided by the sum of the number of mole of all component so for the mole of ethylene glycol what we do that mole of ethylene glycol upon mole of ethylene glycol plus mole of water so put all values and get the answer so you will get that 45 .10 upon 62 .10 and then 45 .10 upon 62 .10, okay? plus that is all over value.

05:17 Wait, i'm just partial otherwise you will getting confused.

05:21 This is a division part.

05:23 Okay, and here is a water value.

05:27 250 divided by 18.

05:30 So from this you get the mole fraction will be 0 .0497.

05:36 Okay? so you will get the 0 .0497...