Concentrations Expression(2)

# Concentrations Expression(2) - CONCENTRATION EXPRESSION 1...

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1 CONCENTRATION EXPRESSION

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2 ) mol / g ( weight formula grams = Moles Where formula weight represents the atomic or molecular weight of the substance. Thus, mol / g 870 . 107 g = wt f g = + Ag Moles mol / g 04 . 142 g = wt f g = 4 SO 2 Na Moles
3 Calculate the number of grams in one mole of CaSO 4 7H 2 O Solution One mole is the formula weight expressed in grams. The formula weight is 262.25 g/mol 14.11 14 H 176.00 11 O 32.06 S 40.08 Ca

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4 ) mmol / mg ( weight formula milligrams = oles lim Mi g Na 2 SO 4 = moles X f wt = moles X 142.04 g/mol g Ag = moles X f wt = moles X 107.870 g/mol We usually work with millimole quantities, so Miligrams = millimoles X formula weight (mg/mmol) Note that g/mol is the same as mg/mmol, g/L the same as mg/ml, and mol/L the same as mmol/mL.
5 Calculate the number of moles in 500 mg Na 2 WO 4 (sodium tungstate). Solution How many milligrams are in 0.250 mmol Fe 2 O 3 (ferric oxide)? Solution 0.250 mmol x 159.7 mg/mmol = 39.9 mg mol 0.00170 = mol/mmol 0.001 x mg/mmol 293.8 mg 500

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6 Millimoles = molarity X milliliters (or mmol = M X mL) A solution is prepared by dissolving 1.26 g AgNO 3 in a 250 – mL volumetric flask and diluting to volume. Calculate the molarity of the silver nitrate solution. How many millimoles AgNO 3 were dissolved? Solution Then, Millimoles = (0.0297 mmol/mL)(250 mL) = 7.42 mmol mmol/mL) 0.0297 (or L 0.0297mol/ = 0.250L 9g/mol 1.26g/169. = M
7 How many grams per milliliter of NaCI are contained In a 0.250 M solution? Solution 0.250 mol/L = 0.250 mmol/mL 0.250 mmol/mL X 58.4 mg/mmol X 0.001 g/mg = 0.0146 g/ml How many grams Na 2 SO 4 should be weighed out to prepare 500 mL of a 0.100 M solution? Solution 500 mL X 0.100 mmol/mL = 50.0 mmol 50.0 mmol X 142 mg/mmol X 0.001 g/mg = 7.10 g

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8 Expressing concentrations of solution Molarity = moles of solute (mol) volume of solution (L) = mmoles of solute (mmol) volume of solution (mL) Normality (N) = no. of equivalents (eq) volume of solution (L) = no. of equivalents (meq) volume of solution (mL)
9 No. equivalents (eq) = mass (g) equivalent mass (g/eq) = normality (eq/L) x vol (L) No. equivalents (meq) = mass (mg) equivalent mass (mg/meq) = normality (meq/mL) x vol (mL) Equivalent mass = Relative Molec Mass (g/mol) No. of reacting units (eq/mol) Reacting units: H + (acid/base) or e - (redox)

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Concentrations Expression(2) - CONCENTRATION EXPRESSION 1...

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