SOAL
- A solution of common salt in water is prepared by adding 20 kg of salt to 100 kg of water, to make a liquid of density 1323 kg/m3. Calculate the concentration of salt in this solution as a (a) weight fraction, (b) weight/volume fraction, (c) mole fraction, (d) molal concentration.
20 / (100 + 20) = 0.167: % weight / weight = 16.7%
(b) Weight/volume:
A density of 1323kg/m3 means that lm3 of solution weighs 1323kg, but 1323kg of salt solution contains
(20 x 1323 kg of salt) / (100 + 20) = 220.5 kg salt / m3
1 m3 solution contains 220.5 kg salt.
Weight/volume fraction = 220.5 / 1000 = 0.2205
And so weight / volume = 22.1%
c) Moles of water = 100 / 18 = 5.56
Moles of salt = 20 / 58.5 = 0.34
Mole fraction of salt = 0.34 / (5.56 + 0.34) = 0.058
d) The molar concentration (M) is 220.5/58.5 = 3.77 moles in m3
Note that the mole fraction can be approximated by the (moles of salt/moles of water) as the number of moles of water are dominant, that is the mole fraction is close to 0.34 / 5.56 = 0.061. As the solution becomes more dilute, this approximation improves and generally for dilute solutions the mole fraction of solute is a close approximation to the moles of solute / moles of solvent.
In solid / liquid mixtures of all these methods can be used but in solid mixtures the concentrations are normally expressed as simple weight fractions.
With gases, concentrations are primarily measured in weight concentrations per unit volume, or as partial pressures. These can be related through the gas laws. Using the gas law in the form:
pV = nRT
where p is the pressure, V the volume, n the number of moles, T the absolute temperature, and R the gas constant which is equal to 0.08206 m3 atm / mole K, the molar concentration of a gas is then
n / V = p/RT
and the weight concentration is then nM/V where M is the molecular weight of the gas.
The SI unit of pressure is the N/m2 called the Pascal (Pa). As this is of inconvenient size for many purposes, standard atmospheres (atm) are often used as pressure units, the conversion being 1 atm = 1.013 x 105 Pa, or very nearly 1 atm = 100 kPa.
2. If air consists of 77% by weight of nitrogen and 23% by weight of oxygen calculate:
(a) the mean molecular weight of air,
(b) the mole fraction of oxygen,
(c) the concentration of oxygen in mole/m3 and kg/m3 if the total pressure is 1.5 atmospheres and the temperature is 25 oC.
(a) Taking the basis of 100 kg of air: it contains 77/28 moles of N2 and 23/32 moles of O2
Total number of moles = 2.75 + 0.72 = 3.47 moles.
So mean molecular weight of air = 100 / 3.47 = 28.8
Mean molecular weight of air = 28.8
b) The mole fraction of oxygen = 0.72 / (2.75 + 0.72) = 0.72 / 3.47 = 0.21
Mole fraction of oxygen = 0.21
(c) In the gas equation, where n is the number of moles present: the value of R is 0.08206 m3 atm/mole K and at a temperature of 25 oC = 25 + 273 = 298 K, and where V= 1 m 3
pV = nRT
and so, 1.5 x 1 = n x 0.08206 x 298
n = 0.061 mole/m3
weight of air = n x mean molecular weight
= 0.061 x 28.8 = 1.76 kg / m3
and of this 23% is oxygen, so weight of oxygen = 0.23 x 1.76 = 0.4 kg in 1 m3
Concentration of oxygen = 0.4kg/m3
or 0.4 / 32 = 0.013 mole / m3
When a gas is dissolved in a liquid, the mole fraction of the gas in the liquid can be determined by first calculating the number of moles of gas using the gas laws, treating the volume as the volume of the liquid, and then calculating the number of moles of liquid directly.
3. In the carbonation of a soft drink, the total quantity of carbon dioxide required is the equivalent of 3 volumes of gas to one volume of water at 0 oC and atmospheric pressure. Calculate (a) the mass fraction and (b) the mole fraction of the CO2 in the drink, ignoring all components other than CO2 and water.
Basis 1 m3 of water = 1000 kg
Volume of carbon dioxide added = 3 m3
From the gas equation, pV = nRT
1 x 3 = n x 0.08206 x 273
n = 0.134 mole.
Molecular weight of carbon dioxide = 44
And so weight of carbon dioxide added = 0.134 x 44 = 5.9 kg
(a) Mass fraction of carbon dioxide in drink = 5.9 / (1000 + 5.9) = 5.9 x 10-3
(b) Mole fraction of carbon dioxide in drink = 0.134 / (1000/18 + 0.134) = 2.41 x 10-3
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