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?

1 Answer
Mar 12, 2016

Answer:

#"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.

chemwiki.ucdavis.edu/Wikitexts/University_of_California_Davis/UCD_Chem_002B

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)|)))#