Percent concentration wv concentration mass of solute in
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Percent (%) concentration % (w/v) concentration: mass of solute in grams contained by 100 m. L solution % (w/w) concentration: mass of solute in grams contained in 100 g of solution % (v/v) concentration: volume of solute in m. L in 100 m. L solution
parts per million (ppm) are used to express the concentrations of very dilute solution A solution whose solute concentration is 1 ppm contains 1 g of solute for each million (106) grams of solution or, equivalently, 1 mg of solute per kilogram of solution A 2. 5 -g sample of groundwater was found to contain 5. 4 g of Zn 2+. What is the concentration of Zn 2+ in parts per million?
Mole Fraction, Molarity, and Molality are concentration expressions based on the number of moles of one or more components of the solution The symbol X is used for mole fraction, fraction with a subscript to indicate the component of interest. The mole fraction of HCl is represented as XHCl. The sum of the mole fractions of all components of a solution must equal 1 A solution of hydrochloric acid contains 36 percent HCl by mass. Calculate the mole fraction of HCl in the solution.
Molarity of a solute in a solution is defined as the number of moles of solute per liters of solution A 2. 00 L solution of hydrochloric acid contains 36 g of HCl. Calculate the molarity of HCl in the solution. 0. 99 mol = 0. 495 = 0. 50 M MHCl = --------2. 00 L
The molality of a solution, denoted m, is defined as the number of moles of solute per kilogram of solvent A solution of hydrochloric acid contains 36 percent HCl by mass. Calculate the molality of HCl in the solution. (c) What additional information would you need to calculate the molarity of the solution? Mass of water = 100 g. Solu – 36 g HCl = 64 g
(a) Calculate the mole fraction of Na. OCl in a commercial bleach solution containing 3. 62 mass percent Na. OCl in water. =0. 00900
(b) What is the molality of a solution made by dissolving 36. 5 g of naphthalene, C 10 H 8, in 420 g of toluene, C 7 H 8? (b) 0. 678 m =0. 28477 /. 420 0. 678024
Given that the density of a solution of 5. 0 g of toluene and 225 g of benzene is 0. 876 g/m. L, calculate (a) the molarity of the solution; (b) the mass percentage of solute.
Calculate the molality of a solution that contains 25 g of H 2 SO 4 dissolved in 80. g of H 2 O
Calculate the molality of a 10. 0% H 3 PO 4 solution in water
Calculate the molality of a solution that contains 51. 2 g of naphthalene, C 10 H—, in 500. m. L of carbon tetrachloride. The density of CCl 4 is 1. 60 g/m. L
If 8. 32 grams of methanol, CH 3 OH, are dissolved in 10. 3 grams of water, what is the mole fraction of methanol in the solution?
What is the molarity of 2500. m. L of a solution that contains 160. grams of NH 4 NO 3?
11 -1 Acid and base Reactions 100. m. L of 0. 100 M HCl solution and 100. m. L of 0. 100 M Na. OH are mixed. What is the molarity of the salt in the resulting solution? Assume that the volumes 1. Balance Chemical Equation are additives. HCl + Na. OH H 2 O + Na. Cl Reaction Ratio: 1 mol 0 10. 0 m mol Start: 10. 0 -10. 0 Change: -10. 0 After reaction: 0. 00 10. 0 For HCl: (100 m. L)(0. 100 mol/L) = 10. 0 mmol HCl For Na. OH = (100. m. L)(0. 100 mol/L) = 10. 0 mmol For Na. Cl: moles of Na. Cl at the end of reaction = moles of HCl at beginning = 10. 0 m mol 10. 0 mmol Na. Cl MNa. Cl = -------------- = 0. 0500 (100+100) m. L solution
11 -1 Acid and base Reactions 80. m. L of 0. 100 M HCl solution and 100. m. L of 0. 100 M Na. OH are mixed. What is the molarity of the salt in the resulting solution? Assume that the volumes are 1. Balance Chemical Equation additives. HCl + Na. OH H 2 O + Na. Cl Reaction Ratio: 1 mol Start: 8. 0 m mol 10. 0 m mol 0 8. 0 m mol Change: -8. 0 After reaction: 0. 00 2. 0 8. 0 For HCl: (80 m. L)(0. 100 mol/L) = 8. 0 mmol HCl For Na. OH = (100. m. L)(0. 100 mol/L) = 10. 0 mmol For Na. Cl: moles of Na. Cl at the end of reaction = moles of HCl at beginning = 8. 0 m mol 8. 0 mmol Na. Cl MNa. Cl = -------------- = 0. 044 (100+80) m. L sol 2. 0 mmol Na. OH MNa. OH = ------------ = 0. 011 180 m. L solution
100. m. L of 0. 100 M HCl solution and 100. m. L of 0. 80 M Na. OH are mixed. What is the molarity of the salt in the resulting solution? Assume that the volumes are 1. Balance Chemical Equation additives. HCl + Na. OH H 2 O + Na. Cl Reaction Ratio: 1 mol 10. 0 m mol 8. 0 m mol 0 Start: Change: -8. 0 +8. 0 -8. 0 After reaction: 2. 0 0. 0 8. 0 For HCl: (100 m. L)(0. 100 mol/L) = 10. 0 mmol HCl For Na. OH = (100. m. L)(0. 80 mol/L) = 8. 0 mmol For Na. Cl: moles of Na. Cl at the end of reaction = moles of Na. OH at beginning = 8. 0 m mol 8. 0 mmol Na. Cl MNa. Cl = -------------- = 0. 040 (100+100) m. L solution 2. 0 mmol Na. Cl MHCl = -------------- = 0. 010 (100+100) m. L solution
100. m. L of 0. 100 M HCl solution and 100. m. L of 0. 80 M Na. OH are mixed. What is the molarity of the salt in the resulting solution? Assume that the volumes are 1. Balance Chemical Equation additives. HCl + Na. OH H 2 O + Na. Cl Reaction Ratio: 1 mol 10. 0 m mol 80. m mol 0 Start: Change: -10. 0 +10. 0 -10. 0 After reaction: 0. 00 70. 8. 0 For HCl: (100 m. L)(0. 100 mol/L) = 10. 0 mmol HCl For Na. OH = (100. m. L)(0. 80 mol/L) = 80. mmol For Na. Cl: moles of Na. Cl at the end of reaction = moles of Na. OH at beginning = 10. m mol 10. mmol Na. Cl MNa. Cl = -------------- = 0. 050 (100+100) m. L solution 70. mmol Na. Cl MNa. OH = -------------- = 0. 035 (100+100) m. L solution
100. m. L of 1. 00 M H 2 SO 4 solution is mixed with 200. m. L of 1. 00 M KOH. What is the molarity of the salt in the resulting solution? Assume that the volumes are 1. Balance Chemical Equation additives. H 2 SO 4 + 2 KOH 2 H 2 O + K 2 SO 4 Reaction Ratio: 1 mol 2 mol 1 mol Start: 100 m mol 200 m mol 0 100 m mol Change: -200 -100 After reaction: 0. 00 0. 0 100 Moles of H 2 SO 4: (100 m. L)(1. 00 mol/L) = 100 mmol mols KOH = (200. m. L)(1. 00 mol/L) = 200 mmol moles of K 2 SO 4 at the end of reaction = moles of H 2 SO 4 at beginning = 100 m mol 100 mmol K 2 SO 4 M H 2 SO 4 = ----------= 0. 333 (100+200) m. L sol
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