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Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol whichshould be added to 4 kg of water to prevent it from freezing at –6°C will be: (Kf forwater = 1.86 K kg mol–1, and molar mass of ethylene glycol= 62gmol–1

Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol whichshould be added to 4 kg of water to prevent it from freezing at –6°C will be: (Kf forwater = 1.86 K kg mol–1, and molar mass of ethylene glycol= 62gmol–1

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4 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
9 years ago
Let the mass of ethylene glycol be xgm
we have freezing point of water as 0 degreeC
depression in freezing in freezing point=0-(-6)=6degreeC
Now depression in freezing point=Kf*m
m is the molality of the solution
m=(x/62)/4=x/248
we have
6=(1.86*x)/248
=800gm
Aniket Tripathi
11 Points
6 years ago
W=6×4000×62÷1000÷1.86=8006=∆t4000=mass of water in gms 62= molecular mass of ethylene glycol1.86= k
Krish Gupta
askIITians Faculty 82 Points
3 years ago
gm,xLet the mass of ethylene glycol be
is the molality of the solutionm=(x/62)/4=x/248we have6=(1.86*x)/248=800gmmCNow depression in freezing point=Kf*m0 we have freezing point of water as 0 degreeCdepression in freezing in freezing point=0-(-6)=6
Hope it helps.
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the answer to your question.
 
Let the mass of ethylene glycol be xgm
we have freezing point of water as 0oC
depression in freezing in freezing point=0 – ( – 6) = 6oC
Now depression in freezing point = Kf*m
m is the molality of the solution
m = (x/62)/4 = x/248
we have
6 = (1.86*x)/248
Hence, x = 248*6/1.86 = 800gm
 
Thanks and regards,
Kushagra

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