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18 g of glucose, C_(6)H_(12)O_(6) is dissolved in 1 kg of water in a saucepan. At what temperature wil the water boil at 1.013 bar pressure ? K_(b) for water is 0.52 K kg mol^(-1). |
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Answer» Solution :Here, we are given `w_(2)=18 G, w_(1)=1 kg = 1000 g, K_(B)=0.52 "kg mol"^(-1)` `M_(2) ("for glucose", C_(6)H_(12)O_(6)) = 72+12+96=180 g mol^(-1)` `Delta T_(b)=(1000 K_(b)w_(2))/(w_(1)M_(2))=(1000" g kg"^(-1)xx0.52"K kg mol"^(-1)xx18 g)/(1000 g xx 180 " g mol"^(-1))=0.52 K` As water BOILS at 37.3 15 K at 1.013 bat PRESSURE, therefore, boiling point of solution `= 373.15+0.052 K = 373.202 K`. |
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