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CBSE Class 12-science Answered

18g glucose is added to 178.2g water the vapour pressure of water for this aqueous solution at 100​​​​​​​​calsias is   
Asked by araj02582 | 08 Apr, 2019, 12:56: PM
answered-by-expert Expert Answer
Given:
 
Weight of glucose = 18 g
 
Vapour pressure of water = 760 torr
 
Weight of water = 178.2 g
 
Moles of glucose =
 
 18 over 180 space equals space 0.1 space mole
 
Moles of water =
 
fraction numerator 178.2 over denominator 18 space end fraction equals space 9.9 space mol
 
Total no. of moles = 0.1 + 9.9 
 
                            = 10 mol
 
Mole fraction of glucose = Change in pressure(ΔP) with respect to initial pressure (P°)
 
fraction numerator increment straight P over denominator increment straight P to the power of degree end fraction equals space fraction numerator 0.1 over denominator 10 end fraction increment straight P space equals space 0.01 cross times 760 space space space space space space space equals space 7.6 space torr Vapour space pressure space of space the space solution space equals space 760 space minus space space 7.6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 752.4 space torr
 
Vapour pressure of the solution is 752.4 torr.
Answered by Varsha | 08 Apr, 2019, 04:50: PM

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