Question

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?

Answer 1

`"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^@"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 `->` 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_"solute"` - the molarity of the solution
`R` - the universal gas constant, usually given as `0.0821("atm" * "L")/("mol" * "K")`
`T` - the absolute temperature

All you have to do here is rearrange this equation to solve for `c_"solute"`, the molarity of a glucose solution that would have an osmotic pressure equal to `"7.7 atm"` at `25^@"C"`.

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

`color(blue)(|bar(ul(color(white)(a/a)T["K"] = t[""^@"C"] + 273.15color(white)(a/a)|)))`

You will have

`c_"glucose" = Pi/(i * RT)`

`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_"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)|)))`