Question: Can someone please explain how he got 0.8 kg? Like what is the step by step algebra to get that answer? Ethylene glycol is used

Can someone please explain how he got 0.8 kg? Like what is the step by step algebra to get that answer?

Ethylene glycol is used as an antifreeze in a cold cliamate Mass of ethylene glycol which should be added to 4kg for water to prevent it from freezing at 6C will be ( Kf for water =1.86Kkgmol1 and molar mass of ethylene glycol =62gmol1 ) =0(6)6cKf=1.86kkgmolltP=Kforolality=kf(wt.ofsolunt(kg)molisofsolente)=1.866241=0.8kg8100

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