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A solution of glucose (Molar mass = 180 g "mol"^(-1) ) in water has a boiling point of 100.20^@C. Calculate the freezing point of the same solution. Molal constants for water K_f and K_b are 1.86 K kg "mol"^(-1) and 0.512 K kg "mol"^(-1)respectively. |
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Answer» Solution :Given that the boiling point `(T_b)` of GLUCOSE solution = `100.20^@C` APPLY the RELATION : ` Delta T_b = K_b xx m` `m = (Delta T_b)/(K_b) = (0.20)/(0.512)` m = 0.390 mol/kg Applying it to the freezing point ` Delta T_f = K_f xx m` ` Delta T_f = 1.86 K kg "mol"^(-1) xx 0.390 "mol" kg^(-1) = 0.725 K` Freezing point of the solution = 273.15 K - 0.725 K = 272.425 K |
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