find how many grams  of KCL  should be added to 1 kg of water to lower its freezing point to         -8 degree celsius( kf =1.86 K Kg/ mol)    pls explain each and every step 

Asked by ppratim02 | 10th Jul, 2016, 07:27: PM

Expert Answer:

KCl dissociate in water completely hence i=2  

 ∆Tf = i Kf x m

m= ∆Tf/i Kf                    

 m= 8 / 2X1.86 = 2.15mol/kg.

Grams of KCl= 2.15 X 74.5 = 160.2 g/kg ( Molar mass of KCl = 74.5g/mol)

Answered by Vaibhav Chavan | 10th Jul, 2016, 09:21: PM