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Unformatted text preview: denominator, consider a solution that just so happens to have 100 mols TOTAL of solute and solvent (that is, choose the size to fit the unit given): mols sucrose mol water mol sucrose mol sucrose mol 100 025 . + = or, the number of mols of sucrose is 2.5. We also then know that 5 . 97 5 . 2 100 == mols of water. Convert 97.5 mols of water to kg: kg g kg mol g mol 76 . 1 1000 1 18 5 . 97 = We now can compute the required molality: m kg mol kg in water mass sucrose mol 42 . 1 76 . 1 5 . 2 = = or 1.42 molal....
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This note was uploaded on 03/30/2008 for the course CH 201 taught by Professor Warren during the Fall '07 term at N.C. State.
 Fall '07
 Warren
 Chemistry, Mole

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