Answer to Question #95183 in General Chemistry for mary

Question

The concentration of glucose (C6H12O6) in normal blood is approximately 94 mg per 100 mL. What is the molarity of the glucose?

Answer 1

Molarity is the number of moles of solute per 1 liter of the solution. At the first let's find the molar mass of glucose - it can be calculated as the sum of atomic weights taking into account the number of atoms of each type in the molecule

$${M_{{C_6}{H_{12}}{O_6}}} = 6{M_C} + 12{M_H} + 6{M_O}$$

Thus

$${M_{{C_6}{H_{12}}{O_6}}} = 6 \cdot 12.01[\frac{{\rm{g}}}{{{\rm{mol}}}}] + 12 \cdot 1.01[\frac{{\rm{g}}}{{{\rm{mol}}}}] + 6 \cdot 16.00[\frac{{\rm{g}}}{{{\rm{mol}}}}] = 180.18[\frac{{\rm{g}}}{{{\rm{mol}}}}]$$

$$\begin{align*} \end{align*}$$

To find molarity we need to divide the number of moles of solute by the volume of solution and the number of moles can be found as division of mass and molar mass of solute

$$M = \frac{\nu }{V} = \frac{{{m_{{C_6}{H_{12}}{O_6}}}}}{{{M_{{C_6}{H_{12}}{O_6}}} \cdot V}}$$

In our case

$$M = \frac{{94[{\rm{mg]}}}}{{180.18[\frac{{\rm{g}}}{{{\rm{mol}}}}] \cdot 100[{\rm{mL]}}}} = \frac{{94 \cdot {{10}^{ - 3}}[{\rm{g]}}}}{{180.18[\frac{{\rm{g}}}{{{\rm{mol}}}}] \cdot 0.1[{\rm{L]}}}} \approx 0.005[\frac{{{\rm{mol}}}}{{\rm{L}}}]$$