A 45.0 g sample of ethyleneglycol, HOCH 2 CH 2 OH, is dissolvedin 590.0 g of wat

Question

A 45.0 g sample of ethyleneglycol, HOCH2CH2OH, is dissolvedin 590.0 g of water. Thevapor pressure of water at 32°C is 35.7 mm Hg. What is thevapor pressure of the water-ethylene glycol solution at 32°C?(ethylene glycol is nonvolatile.)
and answer in mmhg

also i have a problemwith
Some ethylene glycol,HOCH2CH2OH, is added to your car's coolingsystem along with 4.4 kg ofwater. If the freezing point of the water-glycol solution is-15.0°C, how many grams of HOCH2CH2OHmust have been added?
and answer in mmhg

also i have a problemwith
Some ethylene glycol,HOCH2CH2OH, is added to your car's coolingsystem along with 4.4 kg ofwater. If the freezing point of the water-glycol solution is-15.0°C, how many grams of HOCH2CH2OHmust have been added?

Explanation / Answer

For the first question, you should know that for anon-volatile solute, the vapor pressure of the solution is computedwith the following relation: Psolution= X(water)P(water) n(water)=mass(water)/Molar Mass (water)             =590.0/18.00=32.78 mol n(glycol)=mass(glycol)/Molar Mass(glycol)             =45.0/62.0= 0.73mol X(water)=n(water)/n(total)=32.78/(32.78+0.73)=0.98 Psolution=0.98*35.7 ˜ 35mmHg. (Although there is no big difference between the vaporpressures, I think the answer makes sense, since the quantity ofthe solute is so small.) For the second question, the difference intemperature (T) is clearly 15 dregrees Celcius. Tf = Kf.(n(glycol)/m(water)) n(glycol) = (Tf*m(water))/Kf =(15*4.4)/1.86                   = 35.5mol m(glycol)= n(glycol)*M(glycol)=35.5*62= 2201 g = 2.201Kg. I hope this is what you wanted.

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