CBSE Class 11-science Answered
solve the numerical below
Asked by adityalearn28 | 28 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 | 28 Sep, 2010, 11:12: PM
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