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Chp2_chgconc

# Chp2_chgconc - denominator consider a solution that just so...

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An Example of Changing Concentration Units Problem 2.21 of text What is the molality of a sucrose solution if the mole fraction is 0.025? Assume an aqueous solution (water as solvent); Sucrose is a non-electrolyte so i = 1. Step 1: “un-pack” the units. water mol sucrose mol sucrose mol + = 025 . 0 molality = kg in water mass sucrose mol We need mol sucrose then and mol water, which we then convert to mass. The equation on the left, however, is one equation with two unknowns. We will thus use the freedom to consider any “sized” sample. Since the mol fraction has the total # mols in the
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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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