Problem 121 - Chemistry An Atoms First Approach

Question

You make \(20.0 \mathrm{g}\) of a sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) and NaCl mixture and dissolve it in \(1.00 \mathrm{kg}\) water. The freezing point of this solution is found to be \(-0.426^{\circ} \mathrm{C}\). Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.

Step-by-Step Solution

Answer

The mass percent composition of the original sucrose and NaCl mixture is approximately \(40.5\%\) sucrose, and the mole fraction of sucrose in the original mixture is \(0.1034\).

Step 1: Determine the molality using freezing point depression

The freezing point depression, ΔTf, is given by: \[ΔT_f = K_f × m\] where ΔTf is the change in freezing point, Kf is the cryoscopic constant of water, and m is the molality of the solution. We know the freezing point of the solution is -0.426°C, and the freezing point of pure water is 0°C, so the change in freezing point is: \[ΔT_f = 0 - (-0.426) = 0.426°C\] The cryoscopic constant, Kf, for water is 1.86 °C/m (given). Now, we can find the molality of the solution: \[m = \frac{ΔT_f}{K_f} = \frac{0.426}{1.86} = 0.2289\, mol/kg\]

Step 2: Calculate the moles of sucrose and NaCl in the solution

Let x be the molality of sucrose and y be the molality of NaCl, so: \[x + y = 0.2289\] We can set up another equation using the fact that the total mass of the solute mixture is 20.0 g: \[342x + 58y = 20\] where 342 g/mol is the molar mass of sucrose (C12H22O11) and 58 g/mol is the molar mass of NaCl.

Step 3: Solve the system of equations for x and y

We need to solve the following system of equations: \[ \begin{cases} x + y = 0.2289 \\ 342x + 58y = 20 \end{cases} \] Multiplying the first equation by 58, we get: \[58x + 58y = 13.2752\] Now subtract this equation from the second equation: \[(342x + 58y) - (58x + 58y) = 20 - 13.2752\] \[284x = 6.7248\] Dividing both sides by 284, we find: \[x = 0.02368 \, mol/kg\] Now, substituting x back into the first equation: \[0.02368 + y = 0.2289\] \[y = 0.2052 \, mol/kg\]

Step 4: Calculate the mass percent composition and mole fraction of sucrose

Now we can find the mass of sucrose and NaCl in the mixture: Mass of sucrose = molality of sucrose × molar mass of sucrose × mass of water \[Mass\_Sucrose = 0.02368 × 342 × 1 = 8.098 \, g\] Mass of NaCl = molality of NaCl × molar mass of NaCl × mass of water \[Mass\_NaCl = 0.2052 × 58 × 1 = 11.902 \, g\] Now, we can calculate the mass percent composition of sucrose: \[%_{Sucrose} = \frac{Mass\_Sucrose}{Mass\_Sucrose + Mass\_NaCl} × 100\] \[%_{Sucrose} = \frac{8.098}{8.098 + 11.902} × 100 ≈ 40.5\%\] Finally, we can calculate the mole fraction of sucrose in the original mixture: Moles of sucrose (n_sucrose) = molality of sucrose × mass of water \[n_{sucrose} = 0.02368 \, mol\] Moles of NaCl (n_NaCl) = molality of NaCl × mass of water \[n_{NaCl} = 0.2052 \, mol\] Mole fraction of sucrose (χ_sucrose) = n_sucrose / (n_sucrose + n_NaCl) \[χ_{sucrose} = \frac{0.02368}{0.02368 + 0.2052} ≈ 0.1034\] So, the mass percent composition of the original sucrose and NaCl mixture is approximately 40.5% sucrose and the mole fraction of sucrose in the original mixture is 0.1034.

Key Concepts

Molality

Molality is a measure of the concentration of a solute in a solution. It is specifically defined as the number of moles of solute per kilogram of solvent. This concept is particularly useful in scenarios where temperature changes are involved since it does not vary with temperature, unlike molarity where volume can change. In the context of freezing point depression, molality becomes crucial because it directly affects the freezing point change of the solution. To calculate the molality, you can use the formula:\[ m = \frac{moles \ of \ solute}{mass \ of \ solvent \ in \ kg} \]In this particular exercise, molality was determined as part of the calculation adjusting for the freezing point depression. The key is knowing that as solute concentration increases, it tends to lower the freezing point of the solvent, which is water in this case. Hence, the calculated molality helps us to distinguish the contribution of each solute (sucrose and NaCl) present in the solution.

Mass Percent Composition

Mass percent composition is a way to express a concentration of a component in a mixture, often represented as a percentage. It answers the question of what fraction of the total mass of the mixture is due to one particular component. For this calculation, you simply divide the mass of the component by the total mass of the mixture, then multiply by 100.The formula for mass percent composition is:\[ \%_{component} = \left( \frac{mass \ of \ component}{total \ mass \ of \ mixture} \right) \times 100 \]In the exercise, the mass of sucrose and NaCl was found by using their respective molalities, molar masses, and the mass of water. Once those values were established, calculating the mass percentage of sucrose was straightforward. These values tell us that out of the total solute mass, a certain percentage is made up by sucrose, which was about 40.5% in this example.

Mole Fraction

The mole fraction is another way to state the composition of a mixture, representing the ratio of the moles of one component to the total moles of all components present. It provides a picture of the component's relative quantity without invoking mass or volume, making it dimensionless.To determine the mole fraction, use the formula:\[ \chi_{component} = \frac{moles \ of \ component}{total \ moles \ of \ all \ components} \]In the step-by-step solution, both sucrose and NaCl's molalities were turned into moles by considering the mass of water they were dissolved in. The mole fraction offers insight into the proportion of sucrose compared to the total amount of substance present in the water. In this case, it was calculated to be approximately 0.1034, illustrating that sucrose makes up a relatively smaller portion of the mixture compared to NaCl.