Mole-volume calculations

The volume calculated above, 22.4 liters, we have seen before. In Table l-II the pressure-volume product of 32.0 grams of oxygen, O, was found to be 22.4 at 0°C. (Notice that 32.0 grams of Os is the weight of one mole of oxygen.) So we can use this relation,... 

Measurements include initial and final masses, and initial and final volumes. Calculations may include the difference between the initial and final values. Using the formula mass and the mass in grams, moles may be calculated. Moles may also be calculated from the volume of a solution and its molarity. 

The (able below contains solubility and density data lor the salts Na SQ4 and MgS()4. Express their solubilities in terms of molar concentrations, molalities and mole fractions. Calculate the contractions in volume that occur when the solutions are made from the solid salts and the solvent. Comment on the results in terms of the ell ect of ionic charges. The concentrations have been cho-.cn to be comparable. 

Analysis of the equilibrium mixture that results from heating 0.25 mole of C02 to a temperature of 1100°C at a pressure of 10 atm shows the presence of 1.40 x 10 3% 02 by volume. Calculate the value of for the dissociation reaction... 

An equilibrium mixture results from heating 2.00 moles of NOC1 to a temperature of 225.0°C at a pressure of 0.200 atm. Analysis shows the presence of 34.0% NO by volume. Calculate the value of Kp for the dissociation reaction 2NOC1 2NO + Cl at this temperature. 

For the denaturation of / -Lg in 40% 2-chloroethanol, the presently reported partial specific volume measurements result in a AV value of —590 ml/mole. This value is very close to that previously reported for the denaturation of this protein by 6.4M urea, —610 ml/mole (27). Although this similarity of AV values is striking, it might be the result of a fortuitous compensation of various effects. The change in volume calculated from the difference in partial specific volumes is the sum of a number of contributions (28), such as differences in electrostriction in the two media, changes in the density of solvent components when they interact... 

This problem starts out as an application of mole volume. If we had 100 moles of the mixture, then we would have 84 mol CH4, 10 mol ( 2 Ip, 3molC3Hs, and 3 mol N2. We can calculate the amount of natural gas in 100 moles of the mixture by utilizing the molecular weights. 

From pure reagent Select the desired concentration and volume determine the required of moles, then calculate the mass (using the molar mass). 

A titration is a controlled reaction used to determine the number of moles of one substance by treatment with a known number of moles of a second substance. The known number of moles is calculated as a volume times a molarity of the solution of known concentration. The... 

Two tanks are initially sealed off from one another by means Of valve A. Tank I initially contains 1.00 fF of air at 100 psia and 150°F, Tank II initially contains a nitrogen-oxygen mixture containing 95 mole % nitrogen at 200 psia and 200 F. Valve A is then opened allowing the contents of the two tanks to mix. After complete mixing had been effected, the gas was found to contain 85 mole % nitrogen. Calculate the volume of tank n. See Fig. P3.16. 

We ate given thtee of the font ptopetties of ideal gases (moles, volume, and tempetatute), and we ate asked to calculate the fourth (ptessute). Thetefote, we use the ideal gas equation to solve this problem. 

There is an alternative way to do this problem without using the ideal gas law. By definition, one mole of any gas occupies 22.4 L at STP, the standard molar volume of gases. So, if 1.00 mole of C02 occupies 22.4 L, 2.50 moles would occupy 2.50 moles x 22.4 L/mole = 56.0 L, the identical volume calculated with the ideal gas equation. 

The molar volume is 22.4 L at STP. If we know the number of moles, we can immediately calculate the volume at STP. The number of moles is calculated from the mass with the molar mass ... 

Because the excess volume for this system was expected to be small (10), and any effect is second order (15), only a few determinations were deemed necessary. A standard pycnometric technique was used (16). The densities at 29°C of three samples of known mole fraction, in addition to the pure components, were found. The excess volumes calculated from these results are shown as Table I. For the accuracy required by the calculations, a value for VE of 1.0 cm3 mol 1, independent of composition over the range of x considered, was sufficient. The effect of error in VE is discussed later. Differentiation of Equation 12 then leads to... 

Calculate the mass of the empty bottle. Calculate the mass of the collected gas. Use the volume of gas, water temperature, and barometric pressure along with the ideal gas law to calculate the number of moles of gas collected. Use the mass of gas and the number of moles to calculate the molar mass of the gas. 

To calculate the masses of the liquids injected it is necessary to know the densities of the liquids at the temperature and pressure in the injectors. The injectors are not thermostatted but set above the bath in an air-conditioned room. Because the injectors themselves have a large heat capacity, rapid fluctuations in the room temperature will not have a major effect. Long-term drifts in the temperature would be more serious. Compression of the liquids to 500 kPa will cause an increase in density of about 0.05 per cent. Van Ness points out that there will be a negligible error in the mole fraction calculated from atmospheric data provided all volume measurements on both liquids are made at the same applied pressure and the liquids have similar isothermal compressibilities. When the isothermal compressibilities of the two components differ appreciably a more detailed calculation is necessary to determine the mole fraction. 

Finally, substitute the given quantities along with the number of moles just calculated to calculate the volume. 

As another example, consider the effects of a change of pressure, holding both the temperature and the number of moles constant. Calculate the new volume of a quantity of gas occupying initially 16.0 L at a pressure of 0.900 atm when the pressure is changed to 1.20 atm. In this case, both n and T remain the same and cancel out of the equation giving the following relationship ... 

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