130. A solution of glucose in water is labelled as 10% W/W, what would be the molality and mole fraction of each component in the solution? If the density of the solution is 1.2 g mL^–1, then what shall be the molarity of the solution? 

Consider 100 g of 10% solution of glucose whose conc. is 10% W / W.
Mass of solution = 100 g
Mass of glucose = 10 g
Mass of solvent = 100 – 10 = 90 g
Molar mass of glucose,
C₆H₁₂O₆ = 6 x 12 + 12 x 1 + 6 x 16
= 72 + 12 + 96 = 180 g / mol

Molality = $\frac{\mathrm{Mass} \mathrm{of} \mathrm{solute}}{\mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{solute}}\times \frac{1}{\mathrm{Mass} \mathrm{of} \mathrm{solvent} \mathrm{in} \mathrm{kg}}$
            $= \frac{10}{180\times 0.09 \mathrm{kg}} = 0.617 \mathrm{m}.$

     Moles of ${\mathrm{H}}_2\mathrm{O}, {\mathrm{n}}_{\mathrm{A}} = \frac{90}{18} = 5$
           Moles of glucose, ${\mathrm{n}}_{\mathrm{B}} = \frac{10}{180} = 0.055$
                      $\frac{0.055}{5+0.055} = 0.0108 = .01$
Mole fraction of glucose, ${\mathrm{x}}_{\mathrm{B}} = \frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}+{\mathrm{n}}_{\mathrm{B}}}$
Mole fraction of water
            ${\mathrm{x}}_{\mathrm{A}} = 1-{\mathrm{x}}_{\mathrm{b}} = 1-0.0108 = 0.989 = 0.99$
                       Density of sol = 1.2 g mol L
Therefore, volume of solution
                 $= \frac{\mathrm{Mass}}{\mathrm{Density} } = \frac{100}{1.2} = 83.33 \mathrm{mL}$

                  Molarity  = $\frac{10/180}{83.33/100}=0.67$