solve the numerical below

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