document.write( "Question 1001501: If the density of a 500.0g H2SO4 solution is 1.84 g/cm^3 what are its molarity, mole fraction, and molality?\r
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\n" ); document.write( "\n" ); document.write( "Don't know if you can help me. This is the only site I could find. It's chemistry. Do you know of a chemistry site like this one?
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Algebra.Com's Answer #618775 by KMST(5328)\"\" \"About 
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The website yeahchemistry.com will help you with chemistry questions.
\n" ); document.write( "If the solution was 100% sulfuric acid, you would have 500g sulfuric acid and 0g water.
\n" ); document.write( "When you buy a bottle of sulfuric acid it would have a label with analysis data,
\n" ); document.write( "and it could read \"assay = 98.6%\", or something like that.
\n" ); document.write( "However, you may assume, it is 100%, because 1.84g/mL is the densityof 100% sulfuric acid at \"10%5Eo\" .
\n" ); document.write( "(Warmer acid, and/or less concentrated acid has a lower density).
\n" ); document.write( "Then, \"%28500g%29%2A%281mol%2F%2298+g%22%29=5.1mol\" would give you the number of moles of sulfuric acid.
\n" ); document.write( "You can use the density to find the volume of those 500g solution,
\n" ); document.write( "\"500g%2A%281mL%2F1.84g%29=272mL=0.272L%29\" .
\n" ); document.write( "Then, \"5.1mol%2F%220.272+L%22=18.8\"mol/L is the molarity, often written as \"18.8M\" , and read as \"18.8 molar\".
\n" ); document.write( "Since \"500g=0.5kg\" , \"5.1mol%2F%220.5+kg%22=10.2\"mol/kg is the molality,
\n" ); document.write( "which your instructor may express as \"10.2molal\" or \"10.2m\" .
\n" ); document.write( "The mole fraction would be
\n" ); document.write( "moles acid/{moles acid + moles water) ,
\n" ); document.write( "and since we are assuming 0% water,
\n" ); document.write( "that molar fraction would be \"1.0%22\" .
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