InterviewSolution
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InterviewSolution
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19 e C 170 e dtscolved Sl jf‘ «था2: cos s s e 10 ही शा & Lo Gkwaléer otSaucepa - टैग | पटित्त कक ््न्पहि, हि waler I 052 % kjmol AJ दि et bl o |
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Answer» Moles of glucose = 18 g/ 180 g mol–1 = 0.1 molNumber of kilograms of solvent = 1 kgThus molality of glucose solution = 0.1 mol kg-1For water, change in boiling pointÄTb = Kb × m = 0.52 K kg mol–1× 0.1 mol kg–1= 0.052 KSince water boils at 373.15 K at 1.013 bar pressure, therefore, the boiling point of solution will be 373.15 + 0.052 = 373.202 K. |
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