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# 18 g of glucose, $C_6H_{12}O_6$, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? $K_b$ for water is 0.52 K kg $mol^{–1}$.

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## 1 Answer

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No. of moles of glucose = 18 g/ 180 g $mol^{–1}$ = 0.1 mol
Mass of solvent = 1 kg
Molality of glucose solution = 0.1 mol $kg^{–1}$
For water, change in boiling point
$\Delta T_b=K_b\times m=0.52K kg mol^{-1}\times 0.1mol kg^{-1}$=0.052
Since 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.
Hence (C) is the correct answer
answered Jun 11, 2014

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