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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. (K f = 1.86 K Kg mol –1 ) :- Options 1. 42 mg 2. 42 g 3. 38.71 g 4. 38.71 mg

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) :-
Options
1.  42 mg
2.  42 g
3.  38.71 g
4.  38.71 mg

Grade:12th pass

1 Answers

Chhavi Jain
92 Points
6 years ago
Dear Student ,
 
We know that freezing point of  pure water is 0oC.
Using the formula , ΔTf = Kf m
where ‘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/mol
no.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 oC
Notice 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 mnew
mnew ---> 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 g
this 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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