Question

18 gram of glucose is dissolved in thousand gram of water in a saucepan at what temperature will the solution.

Answer 1

Answer:

(Kp for water =0.52k mol-1 , boiling point of pure water =373.15k)

Explanation:

W1 = weight of solvent (H2O) = 1kg

W2 = weight of solute (C6H12O6) = 18 GM

M2 = molar mass solute (c6h12o6)=180g mol-1

Tb= 37315k

Tb= kb×1000×w2/m2×w1 = 0.52×1000×18/180×1000 = 0.052k

Therefore,

Tb=Tb=Tp

0 052k =Tb_373.15

Tb=373.202k