Calculating osmotic pressure: The fomula for low molecular mass starch is `(C_(6)H_(10)O_(5))_(n)`, where n averages `2.00xx10^(2)`. When `0.798g` of starch is dissolved in `100.0 mL` of water solution, what is the osmotic pressure at `25^(@)C`?
Strategy: Work out the molecular mass of `(C_(6)H_(10)O_(5))_(200)`, and use it to obtain the molarity of the starch solution. Substitute into the formula for the osmotic pressure, `pi=CRT`.

Explanation

Step 1. Calculate the molarity
The molecular mass of `(C_(6)H_(10)O_(5))_(200)` is `200 [6xx12+10xx1+5xx16]=32,400 u`. Therefore, the number of moles in `0.798 g` of starch is
`n("moles")=(mass_("starch"))/(molar mass_("starch"))=(0.798 g "starch")/(32.400 g " starch"//mol " starch")`
`=0.0000246 mol` starch
`=2.46xx10^(-5)` mol starch
The molarity of the solution is
`M=n_(starch)/V_(mL)xx(1000 mL)/L`
`=(2.46xx10^(-5) mol)/(100.0 mL)xx(1000 mL)/L`
`=2.46xx10^(-4) mol L^(-1)` solution
Step 2. Calculate osmotic pressure
The osmotic pressure at `25^(@)C` is
`pi=CRT`
`=(2.46xx10^(-4) mol L^(-1))(0.082 (L atm)/(K mol))(298 K)`
`=6.02xx10^(-3)` atm
`=6.02xx10^(-3) "atm" xx(760 mmHg)/(1 "atm")`
`=4.58 mmHg`