Solutions mole problems

C18-0048. Compute how many moles of acid it takes to change the pH of each buffered solution in Problem... 

Heat capacities may be used as factors in factor-label method solutions to problems. Be aware that there are two units in the denominator, mass (or moles) and temperature change. Thus, to get energy, one must multiply the heat capacity by both mass (or moles) and temperature change. 

The first part of this problem appears in numerous problems involving solutions. Moles are critical to all stoichiometry problems, so you will see this step over and over again. This is so common, that anytime you see a volume and a concentration of a solution, you should prepare to do this step. 

Solution This problem is really no different from the molecular weight determinations of unknown acids that are often conducted in general chemistry lab courses. What is important to recognize is that there is one carboxyl group per molecule or one equivalent per mole. Therefore the molecular weight of the polymer is given by... 

Determination of T y. In the formulation of the phase equilibrium problem presented earlier, component chemical potentials were separated into three terms (1) 0, which expresses the primary temperature dependence, (2) solution mole fractions, which represent the primary composition dependence (ideal entropic contribution), and (3) 1, which accounts for relative mixture nonidealities. Because little data about the experimental properties of solutions exist, Tg is usually evaluated by imposing a model to describe the behavior of the liquid and solid mixtures and estimating model parameters by semiempirical methods or fitting limited segments of the phase diagram. Various solution models used to describe the liquid and solid mixtures are discussed in the following sections, and the behavior of T % is presented. 

Solution This problem starts out in the mass portion of the Mole-Go-Round. You have 22.0 grams of C02. (Notice how a number [22.0], units [gram], and substance [C02] are carefully recorded in each step of the problem.) The next step is to convert to moles by dividing by the molar mass. The molar mass for C02 is 44.0 grams/mole. This gives 0.50 moles of C02 as shown in Figure 6.2. 

Note that vapor-pressure lowering, not vapor pressure, is directly proportional to the mole fraction of solute. (See Problem 15.96) For a solution of a nonvolatile (nonevaporating) solute in a volatile (easily evaporated) solvent, the greater the number of moles of solute present, the smaller is the mole fraction of the solvent. The smaller the mole fraction of the solvent, the lower is the vapor pressure of the solvent. Because the solute does not evaporate, the lower the vapor pressure of the solvent, the lower is the vapor pressure of the solution and the larger is the vapor-pressure lowering. That is, the smaller the value of P, the larger is the value of AP. 

Suppose a particular chemical reaction results in the production of 1.5 x 1023 molecules of hydrogen (H2) and you want to determine the number of moles this represents. The solution to this problem may seem obvious to you, but if not, think again about a dozen. If you had six doughnuts, how many dozen doughnuts would you have Dividing 6 by 12 you would find that you have 0.5 (or 1/2) a dozen doughnuts. We would solve the mole problem the same way ... 

Now, use the factor label method to solve this solution stoichiometry problem, just as you used it to solve other stoichiometry problems. Because you know the concentration of the NaOH solution, first find the number of moles of NaOH involved in the reaction. 

Solution The problem is given in grams and the value from Table 8-2 is in moles, so a conversion factor is used ... 

The simplest type of solute distribution problem occurs when A i moles of a solute are completely dissolved and distributed between two immiscible solvents. The. equilibrium di.sEribution of the solute is determined from the single equilibrium relation... 

Strictly speaking, the concentration measures per volume of solvent (Bunsen), per number of moles in the solution (mole fraction), and per mass of solution (samples) are not linearly related, and hence Henry s law cannot simultaneously be valid for all forms. To illustrate the problem, consider the conversion from mole fraction concentrations Xi to per weight concentrations Ci ... 

For this problem we must find the solution mole fractions, the molality, and the molarity. For molarity, we can assume the solution to be so dilute that its density is 1.00 g/mL. We first find the number of moles of lysozyme and of water. 

Now we consider solubility problems in which the solute is a condensed phase (liquid or solid), but the pressure is high and the temperature is near or above the critical temperature of the solvent in such cases, the solvent is not a liquid but a dense fluid. These situations are important because the solubility (i.e., the solute mole fraction in the fluid x ) can be much larger than its value at low pressures. The solubility increases for two reasons ... 

The scrubber in Problem 23 has been running well for several years. It is now observed that the 2g oudet solute mole fraction in air is 0.002. Suggest at least five reasons for this observation. Surest at least five ways to compensate for this problem in the short term. Evaluate each conpensation method as to its suitability. 

Let s take the example of the mixing of 50 ml of a 3 x 10 mol/L barium chloride solution with 100 ml of a 4.5 x 10 mol/L sodium sulfate solution. The problem is to determine the fraction of barium precipitated. Let s call P the number of moles of precipitated barium sulfate per liter of solution. The mass balance relations on Ba +, 504 together with the solubility-product expression provide a system of three equations in three unknowns [Ba +], [804 ], and P ... 

The general XT E problem involves a multicomponent system of N constituent species for which the independent variables are T, P, N — 1 liquid-phase mole fractions, and N — 1 vapor-phase mole fractions. (Note that Xi = 1 and y = 1, where x, and y, represent liquid and vapor mole fractions respectively.) Thus there are 2N independent variables, and application of the phase rule shows that exactly N of these variables must be fixed to estabhsh the intensive state of the system. This means that once N variables have been specified, the remaining N variables can be determined by siiTUiltaneous solution of the N equihbrium relations ... 

Strategy Part (a) is essentially a stoichiometry problem of the type discussed in Chapter 4. For parts (b) and (c), start by calculating (1) the number of moles of OH added and then (2) the number of moles of H+ or OH- in excess. Finally, calculate (3) [H+] and pH. Remember to use the total volume of the solution at that point... 

Solution The obvious way to solve this problem is to choose a pressure, calculate Oq using the ideal gas law, and then conduct a batch reaction at constant T and P. Equation (7.38) gives the reaction rate. Any reasonable values for n and kfCm. be used. Since there is a change in the number of moles upon reaction, a variable-volume reactor is needed. A straightforward but messy approach uses the methodology of Section 2.6 and solves component balances in terms of the number of moles, Na, Nb, and Nc-... 

The process to reach a quantitative solution to the problem requires working with moles. Thus, we need the relationship linking moles to molarity and volume n — M V We must use the equation In two ways ... 

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