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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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Relative decrease in vapour pressure of an aqueous glucose dilute solution is found to be 0.018.What is the elevation in boiling point? (Given 1 molal aq.glucose solution boils at 100.62$^{\large\circ}C$ at 1 atm.pressure)

$\begin{array}{1 1}(a)\;0.12^{\large\circ}C\\(b)\;0.62^{\large\circ}C\\(c)\;0.26^{\large\circ}C\\(d)\;0.21^{\large\circ}C\end{array}$

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Answer: $ 0.62^{\circ}C$
Given 1 molal aq.glucose solution boils at 100.62$^{\large\circ}C$ at 1 atm.pressure.
The boiling point of water is $100^{\circ}C$ at 1 atmosphere of pressure (sea level).
Given this, we can easily calculate the elevation in boiling point as $T - T_0 = 100.62 - 100 = 0.62^{\circ}C$
answered Feb 14, 2014 by sreemathi.v
edited Jul 15, 2014 by balaji.thirumalai
 

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