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`        50 g of antifreeze (ethylene glycol) is added to 200 g water. What amount of ice will separate out at –9.3°C. (Kf = 1.86 K Kg mol–1) :-Options1.  42 mg2.  42 g3.  38.71 g4.  38.71 mg`
2 years ago

Chhavi Jain
92 Points
```							Dear Student , We know that freezing point of  pure water is 0oC.Using the formula , ΔTf = Kf mwhere ‘m’ is the molality . First we will calculate the value of m --->m = no.of moles of solute / wt. of solvent in kg molar mass of ethylene glycol ( C2H6O2 ) = 62 g/molno.of moles of solute = 50 / 62 = 0.806 moles  Now, m= 0.806 / (200 * 10-3 ) = 4.03 molal So,  ΔTf = Kf m =  1.86 * 4.03 = 7.49 = 7.5 oC ( approx ) ΔTf = Tfo –  Tf  7.5 = 0 – Tf  So, Tf = -7.5 oCNotice that we are cooling the solution to  the temperature -9.3oC  which is lower than -7.5 oC .  This implies that molality of the solution will increase because some of the water turns into ice , thus leaving less water in the solution. The no. of moles of ethylene glycol (solute)  will remain unchanged . So, we can write , ΔTf cool= Kf mnewmnew ---> molality of the solution at this new temp.                                               Tfo – Tf = 0- (-9.3) = 9.3 = 1.86 *  (0.806 / wt. of solvent in Kg)                                wt. of solvent = 0.1612 kg or 161.2 gthis is the amount of water at -9.3 oC . So, we can say that the amount of ice that will seperate out is :                                 mass of ice = 200g -161.2g = 38.8 g  ans : option 3.  Chemistry Consultant Askiitians
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2 years ago
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