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8 Cards in this Set
- Front
- Back
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Molarity (M)
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is the number of moles of a substance per liter of solution. "[H+]" is read as "The concentration of H+". Dependent of temperature.
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Molality (m)
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is the concentration expressed as moles of a substance per kilogram of solvent (not total solution). Independent of temperature
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Electrolyte
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is a substance that dissociates into ions in a solution.
-the more dissociated, the stronger the electrolyte -Strong electrolytes are said to have Formal Concentration (F) as their molarity to emphasize that the substance is really converted into other species in solution |
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A solution with a final volume of 500.0mL was prepared by dissolving 25.00mL of methanol (CH3OH, density = 0.7914g/mL) in chloroform. The solution also has a density of 1.454 g/mL. Calculate the Molarity and Molality of the methanol in the solution.
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a.) ((25.00mL)(0.7914g/mL)/(32.042g/mol))/0.5000L = 1.235M
b.) ((19.78g)/(32.042))/0.7072kg = 0.8729m |
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A 48.0 wy% solution of HBr in water has a density of 1.50g/mL
(a) Find the formal concentration of HBr (b) What mass of solution contains 36.0g of HBr? (c) What volume of solution contains 233 mmol of HBr? (d) How much solution is require to prepare 0.250L of 0.160M HBr? |
(a) (48.0g HBr/ 100g solution)(1.50g solution/mL solution) = (0.720g HBr/mL solution) = (720g HBr/L solution)
Formal Concentration = (720g HBr/L)/(80.912g/mol) =8.90M (b) 36g HBr/ (0.480g HBrg/solution) = 75.0g solution (c) 233 mmol = 0.233 mol 0.233mol/8.90 mol/L (or M) = 0.0262L = 26.2mL (d) Mconc x Vconc = Mdil x Vdil (8.90M) x (xmL) = (0.160M) x (250mL) x = 4.49mL |
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A solution contains 12.6 ppm of dissolved Ca(NO3)2 (which dissociates into Ca(2+) + 2NO3). Find the concentration of NO3 in parts per million
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Each mol of Ca(NO3)2 (FM 164.088) contains 2 mol NO3 (FM 62.005), so the fraction of mass that is nitrate is
(2mol NO3/mol Ca(NO3)2)x((62.005 g NO3/mol NO3)/(164.088g Ca(NO3)2/mol Ca(NO3)2)) = 0.7557 g NO3/ g Ca(NO3)2 If the dissolved Ca(NO3)2 has a concentration of 12.6ppm, the concentration of dissolved NO3 is (o.7557)(12.6ppm) = 9.52 ppm. |
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Ascorbic Acid (Vitamin C) reacts with I3 according to the equation
C6H8O6 + I3 +H2O ===> C6H8O7 + 3I + 2H Starch is used as an indicator in the reaction. The end point is marked by the appearance of a deep blue starch-iodine complex when the first fraction of a drop of unreacted I3 remains in the solution. (a) Calculate for the formula mass of ascorbic acid (b) If 29.41mL of I3 solution are required to react with 0.1970g of pure ascorbic acid, what is the molarity of the I3 solution? (c) A Vitamin C tablet containing ascorbic acid plus inert binder was ground to powder, and 0.4242g was titrated by 31.63mL of I3. Find the weight percent of Ascorbic acid in the tablet. |
(a) Formula Mass = 176.124 g/mol
(b) 0.1970g of ascorbic acid/ 176.124g/mol = 1.1185 mmol Molarity of I3 = 1.1185 mmol/29.41mL = 0.03803M (c) 31.63 mL of I3 = 1.203mmol of I3 = 1.203mmol of ascorbic acid =0.2119g = 49.94% f the tablet |
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A solution of NaOH was standardized by titration of a known quantity of the primary standard, potassium hydrogen phthalate (FM 204.221):
C8H5O4K + NaOH ==> C8H4O4NaK + H2O The NaOH was then used to find the concentration of an unknown solution of H2SO4: H2SO4 + 2NaOH ===> Na2SO4 +2H2O (a) Titration of 0.824g of Potassium Hydrogen phthalate required 38.314g of NaOH solution to reach the end point detected by phenolphthalein indicator. Find the concentration of NaOH (mol NaOH/kg solution) (b) A 10.00mL aliquot of H2SO4 solution required 57.911g of NaOH solution to reach the phenolphthalein end point. Find the molarity of H2SO4 |
(a) (0.824 g acid/176.124g/mol) = 4.0348 mmol. This many mmol of NaOH is contained in 0.038 314 kg of NaOH solution.
Concentration=(4.0348 x 10^-3 mol NaOH/ 0.038314 kg solution) = 0.10531 mol/kg solutions (b) mol NaOH = (0.057911kg)(0.10531 mol/kg) = 6.0986 mmol because 2 mol NaOH react with 1 mol H2SO4 [H2SO4]= (1/2(6.0986 mmol))/10.00 mL = 0.305mmol/mL = 0.305M |
