An automobile antifreeze mixture is made by mixing equal...

An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (EG) and water (W) at 20°C. Calculate the a) volume percent, b) the mass percent, c) the molarity, d) the molality and e) the mole fraction of ethylene glycol in the solution. Densities: ethylene glycol = 1.14 g·mL?¹ water = 1.00 g·mL?¹ Molar Mass: ethylene glycol = 62.07 g/mol water = 18.02 g/mol Assume V = 100.00 mL

Transcript

00:01 Okay, so we have given here subparts for this question and we have to find out concentrations and concentration can be found find out in various ways, right? so first of them is volume percent and we have given that equal volumes are mixed.

00:28 Let me write it again.

00:34 Okay and and that is ethylene glycol we have given this symbol in the question right so ethylene glycol is 50 milliliter and w that is water that is again this will be 50 milliliter because the total volume was 100 milliliter so that's why the equal volume for each of ethylene glycol and water will be 50 milliliter each okay so now therefore volume percent so this will be the formula is volume of solute divided by volume of solution multiplied by 100.

01:29 Okay.

01:32 So this will be volume of eg that is a thylene glycol divided by volume of eg divided by volume of eg plus water.

01:48 So whenever the solution volume we have given here actually that was 100 milliliter but if it is not given then we have to add the solute and solvent volume.

02:01 In this case this will be volume of solution simply.

02:07 So can i put the values? so this will be 50 divided by 100 milliliters.

02:17 For each multiplied by 100 and this will be 50%.

02:25 And now moving on to next.

02:29 So this was the answer for first part.

02:33 Moving on to b part.

02:35 So we have first of all we need to calculate mass.

02:40 We have to calculate mass percentage here and for this we have to calculate mass.

02:45 So we know that density is equals to mass divided by volume or now this can also be written as mass is equals to density multiplied by volume.

02:58 Okay.

03:00 So now for eg, that is thylene glycol.

03:08 So mass will be equal to 1 .14 gram per milliliter multiplied by 50 milliliter.

03:21 This will be equal to, so this will be 57 gram.

03:32 Similarly, for water, mass will be equal to 1 gram per millimeter multiplied by 50 milliliter.

03:45 That will be 50 gram.

03:47 Okay.

03:49 So now we have therefore mass per se.

04:02 So this will be equal to mass of solute divide by mass of solution multiplied by 100.

04:24 So putting the values, mass of solute, that is ethylene glycol is 57 gram.

04:31 Mass of solution so that will be solute plus solvent okay mass of solution will be solute plus solvent so here instead of solution you can put mass of ethylene glycol plus water that is 57 gram plus 50 gram multiplied by 100 solving this we will get so this will be 0 .27%...

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