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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?
(Kb for water = 0.52 K kg mol−1, boiling point of pure water = 373.15 K)
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Solution
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
Kb = 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`
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