<p>Here’s some other tough cookies to crack!!!</p>
<p>If vinegar is 5.0% acetic acid(HC2H3O2, mol. wt=60g/mol) in water, what is the molality of vinegar?</p>
<p>An aqueous solution of sucrose(C12H12O11, mol. wt=342g/mol) is 20% sucrose by mass. Calculate the vapor pressure of the solution at 100C.</p>
<p>Chemistry…lol…</p>
<p>B to the U to the M to the P.</p>
<p>If vinegar is 5.0% acetic acid(HC2H3O2, mol. wt=60g/mol) in water, what is the molality of vinegar?</p>
<ul>
<li>Unless there’s some trick to this problem…</li>
</ul>
<p>Molality = kg solvent/mol solute</p>
<ul>
<li>So if vinegar is 5.0% acetic acid in water, then basically the solution is 5.0 mol of acetic acid for every 95.0 mol of water:</li>
</ul>
<p>5.0mol acetic acid/95.0 mol water</p>
<ul>
<li>Now convert the moles of acetic acid to kg…</li>
</ul>
<p>(5.0mol acetic acid/95.0mol water) x (0.060kg acetic acid/mol) = 0.30kg acetic acid/95.0mol water = 0.0032 kg acetic acid/mol water</p>
<p>An aqueous solution of sucrose(C12H12O11, mol. wt=342g/mol) is 20% sucrose by mass. Calculate the vapor pressure of the solution at 100C.</p>
<p>Vapor pressure = mole fraction of water in the soln x vapor pressure of the pure solvent (Raoult’s Law)</p>
<p>That problem didn’t give you the vapor pressure of water at 100C, though, so I’m not sure what you’re suppose to do.</p>
<p>Assume that you have 1200 g of solution
Thus you have (.05)1200=60 g of vinegar=1 mol of vinegar.
Then you have 1200-60=1140 g of water=1.14 kg
Molality= moles of solute/kg of water
Thus, 1 mol of vinegar/1.14 kg of water=.877 m</p>
<p>Whoops - przfer is right. I think if you do the problem the way I did it but switch some stuff around (because I mixed up molality) it should work out to the same answer. Sorry, I haven’t really done Chem since May 11, 2004. :p</p>