Maple syrup sap is 3.0%sugar (sucrose, C_12H_22O_11 MW = 342.3) and 97.0% water

Question

Maple syrup sap is 3.0%sugar (sucrose, C_12H_22O_11 MW = 342.3) and 97.0% water (MW = 18.0) by mass. Maple syrup is produce by heating the sap to evaporate a certain amount of water. It the finished product boils 4 degree C higher than the original sap, calculate the concentration of sugar in the final product in percent composition. Assume that sugar is the only solute and the operation is done at 1 atm pressure. The boiling point of pure water at 1 atm is 100. degree C and the K_b for water is 0.52 degree C kg/mol.

Explanation / Answer

first we find Boiling point of solution i.e original sap

given it has 3 % sugar and 97 % water

hence per 97 g water we have sucrose 3g

sucrose moles = mass / Molar mass of sucrose

          = 3 g / (342.3 g/mol) = 0.00876424

water mass = 97 g = 0.097 kg

molality of sucorse = moles of sucrose / water mass in kg

               = 0.00876424 / 0.097 = 0.09035

elevation in boiling point dT is given by

dT = i x Kb x m    where i = vantff factor = 1 for non dissociable solute

dT = 1 x 0.52 Kg/mol x 0.09035   = 0.047

hence boling point of solution = 100+0.047 = 100.047

now since some water evaporated solution boiling new = 4 + 100.047 = 104.047 ( since mentioned 4C higher than orginial sap)

now dT = 104.047-100 = 4.047 , we use formula to find molality

4.047 = 1 x 0.52 x m

m = 7.7827   = moles of surcose / water mass in kg

let water mass be 1 kg then surcrose moles = 7.7827

surcrose mass = moles x molar mass of sucrose

      = 7.7827 mol x (342.3 g/mol) = 2664 g

water mas s= 1000 g

solution mass = 2664 + 1000 = 3664 g

sucrose % = ( 100 x surcrose mass / solution mass)

      = ( 100 x 2664 /3664) = 72.7   %

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