A solution containing 62 g ethylene glycol in 250 g water is cooled to - 10 o C . If K f for water is 1.86 K kg mol - 1 , the amount of water (in g ) separated as ice is:

Step-by-step solution

Molecular weight of ethylene glycol C 2 H 6 O 2 = 62 Moles of ethylene glycol = W t M . W t = 62 62 = 1   ∆ T f = i × k f × m ∴ ∆ T f = 0 - - 10 = 10 i = 1 for ethylene glycol ∆ T f = 1 × 86 × 1 250 1000 ∆ T f = 1.86 × 4 = 7.44 As ∆ T f = 10 it implies that some amount of water has frozen and this has led to a greater depression of freezing point. 10 = 1.86 × 62 62 W t   o f   H 2 O 1000 W H 2 O = 186   g m Hence, the weight of wa