Practice Exam 4 C/P Section Passage 5 Question 23 Practice Exam 4 C/P Section Passage 2 Question 7 Practice Exam 3 C/P Section Passage 5 Question 22 Practice Exam 2 C/P Section Passage 1 Question 3 Section Bank C/P Section Passage 9 Question 73 Section Bank C/P Section Passage 4 Question 34 Section Bank C/P Section Passage 3 Question 22 The equation c 1V 1В = c 2V 2, where c is concentration and V is volume can be used to calculate changes in concentration from volume changes.Д¬hemistry Question Pack Passage 4 Question 20 When dilution occurs the volume of a solution is changed and this has an impact on the concentration of a solution. For example, if we wanted to know the concentration (M) of 7 grams of sodium hydroxide (NaOH) in 700 mL of water we could do the following calculation: To determine the molarity of a substance, the number of moles of the object is divided by 1 liter (L) of the solvent. The molarity is important to know when planning out experiments and reactions the larger the molarity of a chemical, the higher the concentration. Then, we divide the number of moles by the total solution volume to get concentration. We do this by dividing by the molecular weight of NaCl (58.4 g/mole). If there are 10.0 grams of NaCl (the solute) dissolved in water (the solvent) to produce 2.0 L of solution, what is the molarity of this solution?ДЇirst, we must convert the mass of NaCl in grams into moles. If the amount of solute is given in grams, we must first calculate the number of moles of solute using the soluteвЂ™s molar mass, then calculate the molarity using the number of moles and total volume. To calculate the molarity of a solution, the number of moles of solute must be divided by the total liters of the solution produced. The unit mol/L can be converted to mol/m 3В using the following equation:ДЎ mol/L = 1 mol/dm 3В = 1 mol dm в€’3В = 1 M = 1000 mol/m 3 A solution that contains 1 mole of solute per 1 liter of solution (1 mol/L) is called вЂњone MolarвЂќ or 1 M. However, mol/L is a more common unit for molarity. The SI unit for molar concentration is mol/m 3. Molar concentration (c i ) is calculated by dividing the moles of solute (n i ) by the total volume (V) of the: Windage and drift are also synonymous.In chemistry, the concentration of a solution is often measured in molarity (M), which is the number of moles of solute per liter of solution. (Note: Draw-off and blowdown are synonymous. In some large power plants, the cooling tower concentration cycles may be much higher. W = about 0.01 percent or less of C if the cooling tower has windage drift eliminatorsД¬oncentration cycles in petroleum refinery cooling towers usually range from 3 to 7. W = 0.1 to 0.3 percent of C for an induced draft cooling tower W = 0.3 to 1.0 percent of C for a natural draft cooling tower Windage losses (W), in the absence of manufacturer's data, may be assumed to be: О”T = water temperature difference from tower top to tower bottom, in вЊД¬ p = specific heat of water = 4.184 kJ / kg / вЊ H V = latent heat of vaporization of water = ca. X C / X M = Cycles = M / (D + W) = M / (M Г» E) = 1 + ДЇrom a simplified heat balance around the cooling tower: M (X M) = D (X C) + W (X C) = X C (D + W) Since the evaporated water (E) has no salts, a chloride balance around the system is: X C = Concentration of chlorides in circulating water (C), in ppmwД¬ycles = Cycles of concentration = X C / X MДЄ water balance around the entire system is: X M = Concentration of chlorides in make-up water (M), in ppmw X = Concentration in ppmw (of any completely soluble salts Г usually chlorides) Here are the governing relationships for the makeup flow rate, the evaporation and windage losses, the draw-off rate, and the concentration cycles in an evaporative cooling tower system:

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