18 g of glucose is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100º C is:

7.60 Torr

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76.00 Torr

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752.40 Torr

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759.00 Torr

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Solution

The correct option is
C

752.40 Torr
As we know that weight of solute (glucose) = 18 grams
Weight of solvent (water) = 178.2 grams
Now Number of moles of solute =
$\frac{18}{180} =$
$M o l e s o f s o l v e n t = \frac{W e i g h t o f w a t e r}{M o l e c u l a r w e i g h t o f w a t e r} = \frac{178.2}{18}$
$VaporPressureofwater=Totalpressure×Molefractionofwater$
$V a p o r p r e s s u r e o f w a t e r = 760 \times \frac{m o l e s o f w a t e r}{T o t a l m o l e s}$
$V a p o r p r e s s u r e = 760 \times \frac{\frac{178.2}{18}}{\frac{178.2}{18} + \frac{18}{180}}$
$V a p o r p r e s s u r e o f w a t e r = 752.4 t o r r$

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18 g glucose $(C_{6} H_{12} O_{6})$ is added to 178.2 g water. The vapor presure of water (in torr) for this aqueous solution is:

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