# The osmotic pressure of blood is 7.7 atm at 25°C. What concentration of glucose, C_6H_12O_6, is isotonic (same osmotic pressure) with blood?

Mar 12, 2016

${\text{0.31 mol L}}^{- 1}$

#### Explanation:

The problem wants you to determine what concentration of glucose would produce a solution that has the same osmotic pressure as blood at ${25}^{\circ} \text{C}$.

As you know, osmotic pressure is defined as the pressure required to prevent the flow of water across a semi-permeable membrane from a region of lower solute concentration into a region of higher solute concentration $\to$ think osmosis.

Osmotic pressure can be calculated using the formula

color(blue)(|bar(ul(color(white)(a/a)Pi = i * c_"solute" * RTcolor(white)(a/a)|)))" ", where

$\Pi$ - the osmotic pressure of the solution
$i$ - the van't Hoff factor, equal to $1$ for non-electrolytes
${c}_{\text{solute}}$ - the molarity of the solution
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature

All you have to do here is rearrange this equation to solve for ${c}_{\text{solute}}$, the molarity of a glucose solution that would have an osmotic pressure equal to $\text{7.7 atm}$ at ${25}^{\circ} \text{C}$.

Make sure to convert the temperature from degrees Celsius to Kelvin by using the conversion factor

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} T \left[\text{K"] = t[""^@"C}\right] + 273.15 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You will have

${c}_{\text{glucose}} = \frac{\Pi}{i \cdot R T}$

c_"glucose" = (7.7 color(red)(cancel(color(black)("atm"))))/(1 * 0.0821( color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25) color(red)(cancel(color(black)("K"))))

${c}_{\text{glucose" = "0.3146 mol L}}^{- 1}$

Rounded to two sig figs, the answer will be

c_"glucose" = color(green)(|bar(ul(color(white)(a/a)"0.31 mol L"^(-1)color(white)(a/a)|)))