A 1.00 kg sample of ethylene glycol (C_2 H_6 O_2) is added to 4450 g of...

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science chemistry

Answer 1

Answer #1

∆Tb = 1000×Kb×w/W×m

W is mass of solvent, w is mass of solute ,m is molecular mass of solute.

∆Tb =(Tf-100) =1000×0.512×1/4.450×62

Tf-100 = 1.98

Tf = 100+1.98 =101.98℃

Answer 2

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