Question

1200 g of antifreeze, C2H6O2, is mixed with 1000 g of water, H2O. What is the boiling point of the solution? Kb.p= 0.51 degrees C/m for water

Answer 1

Step 1,-

Calculate molality (m) of the antifreeze , C2H6O2 = ( number of moles of C2H6O2 / one Kg ( or 1000 gms ) of .....................................................................................water )..................

.......................................................................... = ( weight of of C2H 6O2 / molar mass of C2 H6 O2) / 1Kg of water ..........................................................................= ( 1200 / 62 )..................

............................................................................=19.35 / 1

............................................................................= 19.35 m

Step-2,

Apply the relation , T_b = Kb x m .........where T_b  = T_b - T_b^o , m = molality of solution = 19.35 m

Therefore, T_b - T_b^o = K_b .m

Given T_b o = Boiling point of pure water = 373 K , K_b  = 0.51

Substituting the given values we get,

T_b  - 373 = 0.51 x 19.35 .........(.where T_b  = Boiling point of solution ).......

..........T_b ...= 373 + 9.87

................. = 382.87 K

............or, ..= 109.87^o C

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