If the concentration of glucose $$\left( {{C_6}{H_{12}}{O_6}} \right)$$   in blood is $$0.9\,g\,{L^{ - 1}},$$   what will be the molarity of glucose in blood ?

Solution :
$$\eqalign{ & {\text{Molarity}} \cr & = \frac{{{W_B}\left( {{\text{in}}\,\,g} \right)}}{{{M_B}\left( {{\text{in}}\,\,g\,\,mo{l^{ - 1}}} \right) \times {\text{Volume of solution (in }}L)}} \cr & = \frac{{Conc.\left( {{\text{in}}\,\,g\,\,{L^{ - 1}}} \right)}}{{{M_B}\left( {{\text{in}}\,\,g\,\,mo{l^{ - 1}}} \right)}} \cr & = \frac{{0.9}}{{180}} \cr} $$
$$ = 0.005\,M$$     $$\left( {\because \,\,{\text{Molar mass of}}\,\,{{\text{C}}_6}{H_{12}}{O_6} = 180\,g\,mo{l^{ - 1}}} \right)$$