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InterviewSolution
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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`. |
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Answer» 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` |
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