Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Joules23 on January 26, 2008, 04:15:29 PM
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1) A solution of phosphoric acid was made by dissolving 10.0 g of H3PO4 in 102.8 mL of water. The resulting volume was 110. mL. Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of 1.00 g/cm3.
When calculating the Molality, i dont see what im doing wrong because its coming up as incorrect. This is what im doing:
(10g H3PO4/97.99g H3PO4)/.1028 Kg H2O
m=.993mol H3PO4/Kg H20
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7)Acetic acid is a weak acid that ionizes in solution as follows.
CH3COOH(aq) CH3COO-(aq) + H+(aq)
If the freezing point of a 0.106 m CH3COOH solution is -0.202°C, calculate the percent of the acid that has undergone ionization.
After i get the "i" (1.0245) , what do i do with that to get the percent of ionization?
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m=.993mol H3PO4/Kg H20
Seems OK to me.
Final volume is of no use here, but it should be 108, not 110 mL.
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7)Acetic acid is a weak acid that ionizes in solution as follows.
CH3COOH(aq) CH3COO-(aq) + H+(aq)
If the freezing point of a 0.106 m CH3COOH solution is -0.202°C, calculate the percent of the acid that has undergone ionization.
After i get the "i" (1.0245) , what do i do with that to get the percent of ionization?
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percent ionization = mol acid ionized / mol acid available x 100
if there is 1.0245 mol of particles and only 1 mol (undissociated) thought to be available, how many moles were ionized?
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(1.02/2.00)(100) = 51.0% ?
(1.0245=1.02 using correct sig figs)
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(1.02/2.00)(100) = 51.0% ?
(1.0245=1.02 using correct sig figs)
If the acid were undissociated, then DT measured = DT calculated, i would be = 1
but since i is greater than 1 then some of the acid ionized. How much greater than 1 indicates the mol fraction of acid that was ionized.
so 1.0245 (dissociated) - 1.00 (undissociated acid) = 0.0245 x 100 = percent ionization