Question

The osmotic pressure of human blood is 7.6 atm at 37 °C. What mass of glucose,\({\bf{C}}{}_6{{\bf{H}}_{12}}{{\bf{O}}_6}\), is required to make 1.00 L of aqueous solution for intravenous feeding if the solution must have the same osmotic pressure as blood at body temperature, 37 °C?

Short answer

The Mass of Glucose, ,\({\rm{C}}{}_6{{\rm{H}}_{12}}{{\rm{O}}_6}\), is 54g.

Step-by-step solution

Osmotic Pressure

It may be defined as the pressure exerted on the solvent to prevent it to pass through the osmosis. It is often used to express the concentration of a solution.

\[\Pi {\bf{ = iMRT}}\]

Where,

π = Osmotic pressure

i = Van't Hoff's factor

M = Molarity

R = Ideal gas constant

T- Temperature

Molarity of Glucose

Molarity of glucose by using osmotic pressure,

\(\begin{aligned}{\rm{M}} &= \;\dfrac{{\rm{\Pi }}}{{{\rm{RT}}}}\\{\rm{M}} &= \;\dfrac{{{\rm{7}}{\rm{.6}}\;{\rm{atm}}}}{{\left( {{\rm{0}}{\rm{.08206}}\;{\rm{L}}\;{\rm{atm}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}{{\rm{K}}^{{\rm{ - 1}}}}} \right){\rm{ \times }}\left( {{\rm{310}}\;{\rm{K}}} \right)}}\\ &= \;{\rm{0}}{\rm{.30}}\;{\rm{mol}}\;{{\rm{L}}^{{\rm{ - 1}}}}\end{aligned}\)

\(\begin{aligned}{\rm{Number}}\;{\rm{of}}\;{\rm{moles}} &= \;{\rm{0}}{\rm{.30}}\;{\rm{mol/L \times 1}}{\rm{.00}}\;{\rm{L}}\\{\rm{Number}}\;{\rm{of}}\;{\rm{moles}} &= \;{\rm{0}}{\rm{.30}}\;{\rm{mol}}\\{\rm{Mass}}\;{\rm{of}}\;{\rm{glucose}} &= \;{\rm{180}}{\rm{.157}}\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}{\rm{ \times 0}}{\rm{.30}}\;{\rm{mol}}\\{\rm{Mass}}\;{\rm{of}}\;{\rm{glucose}} &= \;{\rm{54}}\;{\rm{g}}\end{aligned}\)