# 25 ppb diluted 5 ml in 100 ml 500 ppb undiluted 5.00 x 10-5 g in g = %

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1 1. If a solution is determined to be 25 ppb magnesium and this was formed by dissolving g of an unknown solid into 100 ml and then diluting 5 ml of that solution into a 100 ml volumetric flask. What was the concentration of magnesium in the original solid. 25 ppb diluted 5 ml in 100 ml 500 ppb undiluted 5.00 x 10-5 g in g = % 2. If a solution is determined to be 1.56 ppm sodium and this was formed by taking a 5 ml aliquot of river water and diluting to 100 ml. What would the concentration of the water sample be in mg/l? If all the sodium originated from sodium chloride, what would the concentration of NaCl be in mg/l? in mol/l? 1.56 ppm diluted by 5mL in 100 ml ( a 20 times dilution, therefore) 31.2 ppm undiluted or 3.12 x 10-3 g in 100 ml = 31.2 mg/l Na Na g/mol = mol of NaCl g/mol 79.3 mg/l NaCl

2 3. A river water and a tap water sample are analyzed for Nitrate (NO 3 - ) and Nitrite (NO 2 - ) by analyzing a colored complex of Nitrite (NO 2 - ). A series of Nitrate standards were prepared and then converted to Nitrite by passing them through a cadmium column. The concentration and absorbance of the corresponding colored complex of these standards is presented in the table below. The calibration curve below was also generated from this data. Analogous measurements on the river water and tap water are also presented below. Determine the concentration of Nitrate (NO 3 - ) and Nitrite (NO 2 - ) in the samples. Conc. Of NO nm nm River Water Through Cd column River Untreated by Cd Tap Water Through Cd column Tap Water Untreated by Cd Absorbance y = x R 2 = Conc. (mg/l) From Equation of Line: X = (Y-Intercept)/Slope Sample Type Absorption X calc River Water through Cd Column River Water untreated by Cd Tap Water through Cd Column Tap Water untreated by Cd Slope Intercept River NO ppm River NO ppm Tap NO ppm tapno ppm

3 4a) Determine the hydroxide concentration [OH-] in Molarity of a solution of Calcium Hydroxide that made by dissolving solid Calcium Hydroxide into deionized water if the calcium concentration is determined to be 100 ppm. b) what would the ph of this solution be? 100 ppm Ca g/l Ca g/mol Ca mol/l Ca 2 mol OH- per 1 mol Ca in Ca(OH) 2 5.0E-03 mol/l OH 2.30 poh ph 5. A solution is made up by dissolving grams of MgBr 2 and g of KBr in 100 ml of deionized water. If 10 ml of this solution is placed into a column containing an ion exchange resin, how many moles of H + will be displaced? (Hint: one H + is displaced by a +1 charged cation, 2 H + by a +2 charged cation). If the displaced acid is collected in a beaker, how many mls of M NaOH would be required to neutralize the solution? g MgBr g/mol MgBr mol 100 ml 1.17E-02 mol/l MgBr E-02 mol/l Mg g KBr 119 g/mol KBr mol 100 ml 1.68E-02 mol/l K 1.68E-02 mol/l K + 10 ml 1.17E-04 mol Mg +2 1 mol Mg +2 = 2 mol H E-04 mol H + Displaced Mg ml 1.68E-04 mol K + 1 mol K + = 1 mol H E-04 mol H + Displaced K E-04 mol H + Displaced Mg +2 and K mol/l NaOH ml NaOH required

4 6) Typical seawater contains 2.7 grams of salt (sodium chloride, NaCl) for every 100 ml. Given this, what is the molarity of NaCl in the ocean? (6b) MgCl 2 has a concentration of M in the ocean. How many grams of MgCl 2 are present in 25 ml of seawater? SOLUTION The molecular weight of NaCl is (Na) (Cl) = g/mol. The moles of salt in 2.7 g are (2.7 g)/(58.44 g/mol) = mol, so the molarity is; [NaCl] = (0.046 mol)/(0.100) = 0.46 M Note: Significant Figures in 100 ml ambiguous, also could be 0.5 M SOLUTION (6b) The molecular weight of MgCl 2 is ( (35.45)=) g/mol. The number of grams in 25 ml is therefore; mol/liter x g/mol x Liter = g = 0.13 grams 7) Find the Molarity and Molality of 37.0 wt % HCl. The density of this solution is given as 1.19 g/ml. SOLUTION Molarity is the # moles/liter need to find the number of moles Mass of one Liter of solution = 1.19 g/ml x 1000 ml = 1190 g Mass of HCl = 1190 (g Solution)/Liter x (g HCl)/(g Solution) = 440 (g HCl)/Liter Mol. Wt. HCl = g/mol (440 g/liter) (36.46 g/mol) = 12.1 mol/liter = 12.1 M In g of solution there are 37.0 g HCl and 63.0 g of H 2 O (37.0 g (36.46 g/mol)) (0.063 kg H 2 O) = Molality = 16.1 m

5 8) Cupric Sulfate pentahydrate (CuSO 4 5H 2 O) with a molecular weight of g/mol is used to prepare a 500 ml solution of 8.00 mm Cu +2. How many grams of Cupric Sulfate pentahydrate are required? SOLUTION: The number of moles of 8.00 mm Cu +2 in 500 mls are; 8.00 x 10-3 mole/l x L = 4.00 x 10-3 mol Cu +2 required 4.00 x 10-3 mol x g/mol = grams required 9) How many grams of MgCl 2 (95.20 g/mol) are required top make a 500 ml solution of 8.00 mm Cl -? Answer: grams from 2.00 x 10-3 mol of MgCl 2 10) What would be the procedure used to make the solution specified by 8? What would be the procedure used to make the solution specified by 8? Using a clean 500 ml volumetric flask, add g of CuSO 4 5H 2 O followed by ~300 to 400 ml distilled H 2 O. Swirl to dissolve reagent, then dilute to mark. Invert covered flask approximately 20 times to ensure well mixed.

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