QJEEM06Ch2eb5: JEE Main 2006 Chemistry - Class 12 JEEM06Ch2eb5 Solution: 18 g of glucose (C 6 H 12 O 6 ) is added to 178.2 g of water.…

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18 g of glucose (C₆H₁₂O₆) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100^oC is

Accepted Answer

Moles of glucose

= {{18} \over {180}} = 0.1

Moles of water

= {{178.2} \over {18}} = 9.9

Total moles

= 0.1 + 9.9 = 10

{P_{{H_2}O}} =

Mole fraction

\times

Total pressure

= {{9.9} \over {10}} \times 760

= 752.4\,\,

Torr

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