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

An automobile antifreeze mixture is made by mixing equal volume of ethylene glycol and water at 25 oC. The density of the mixture is 1.070 g mL -1 . Given the density and molar mass of pure ethylene glycol are 1.14 g mL -1 and 62.07 g mol -1 respectively, calculate the concentration of ethylene glycol in each of the following. i. Mass percentage. ii. Molarity. iii. Molality.

Transcript

00:01 Hello, in this question we need to find first the value of this mass percent, second, molarity and the third one is that is molality.

00:26 So, first to calculate the concentration or ethylene glycon as the mixture is mixed with equal volumes of ethylene, ethylene.

00:42 Glycol and water at 25 degrees celsius.

00:49 So assuming that there is 10 m l of ethylene glycol, it is reacting with or it is mixed with 10 m l of water.

01:00 They both are mixed.

01:01 So here for ethylene glycol density it is given that is 1 .14 gram per mole and the molar mass of this of ethylene glycol it is 62 .07 gram per mole.

01:20 So therefore here the mass will be as the density is equal to mass divided by volume.

01:27 So therefore here the mass will be density that is 1 .1 4 multiply by volume it is there that is 10 and this will become 11 .4 gram.

01:40 Similarly in case of water volume it is 10m.

01:47 Density it is density of water it is 1 gram per emil and the molar mass it is 18 .01 gram per mole so therefore here the mass will be in this case it is equal to is equal to density multiply by volume and this will become 10 gram now here now as the ethylene glycol and the water they got mixed up that is that means they form the solution.

02:27 They together form the solution.

02:29 And therefore the solution mass will be equal to the mass of these two.

02:34 That means 11 .5 plus 10 it is equal to 21 .4 gram.

02:43 So therefore, and the volume of solution, it is equal to mass divided by density...

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