solve the numerical below

Asked by adityalearn28 | 28th Sep, 2010, 10:43: PM

Expert Answer:

Let us consider 10gm of glucose (solute) present in 100gm of solution.
and, d=m/v (Density=Mass/Volume)
 
1.2=100/V
 
V=100/1.2
 
V=83.33 ml
 
V=83.33/1000=.083L
 and,
 
Molarity=No. Of Moles of solute/Volume Per litre of solution
 
No. Of Moles of solute=wt of solute/Mol.Wt
 
Mol.wt of Glucose(C6H12O6)=180
 
No. Of Moles of solute=10/180
 
 
M=10/180/.08333L
 
=.66M
 
Now, Consider that solvent is H20 which is 90%.
 
No. Of Moles of solute=10/180=.0555
 
No. Of Moles of solvent=90/18=5
 
Mole fraction of glucose=moles of glucose/(Moles o glucose+moles of water)
 
Mole fraction of glucose=.0555/(5+.055)
 
=.0109

Answered by  | 28th Sep, 2010, 11:12: PM

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