Question

8 g of glucose, C₆H₁₂O₆ (Molar Mass = 180 g mol⁻¹) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil? 

(K_b for water = 0.52 K kg mol⁻¹, boiling point of pure water = 373.15 K)

Answer 1

w₁ = weight of solvent (H₂O) = 1 kg and w₂ = weight of solute (C₆H₁₂O₆) = 18 gm

M₂ = Molar mass of solute (C6H12O6) = 180 g mol⁻¹

K_b = 0.52 K Kg mol⁻¹

`T_b^o=373.15k`

`therefore triangle T_b=(K_bxx1000xxw_2)/(M_2xxw_1)=(0.52xx1000xx18)/(180xx1000)=0.052k`

`therefore triangle T_b=T_b-T_b^o=>T_b-373.15=>T_b=373.20K`