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Chapter 11: Problem 121

You make \(20.0 \mathrm{~g}\) of a sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) and \(\mathrm{NaCl}\) 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.

Short Answer

Expert verified
The mass percent composition of the original mixture is 42.2% sucrose and 57.8% NaCl. The mole fraction of sucrose in the original mixture is 0.0697.

Step by step solution

01

Freezing Point Depression Formula

Find the formula for the freezing point depression of a solution. The formula is given by: \[ \Delta T_{f} = K_{f}\cdot m \cdot i \] where \(\Delta T_{f}\) is the freezing point depression, \(K_{f}\) is the cryoscopic constant, \(m\) is the molality of the solution, and \(i\) is the van 't Hoff factor.
02

Cryoscopic Constant of Water

Find the cryoscopic constant, \(K_{f}\), for water. It is given as \(1.86^\circ \mathrm{C\,kg/mol}\).
03

Molality and Freezing Point Depression

Use the given freezing point depression of \(0.426^{\circ} \mathrm{C}\) and the cryoscopic constant (\(1.86^{\circ} \mathrm{C\,kg/mol}\)) to solve for the molality of the solution: \[ m = \frac{\Delta T_f}{K_f \cdot i} = \frac{0.426^{\circ} \mathrm{C}}{1.86^{\circ} \mathrm{C\,kg/mol} \cdot 1} = 0.229 \, \mathrm{mol/kg} \] Note that we used an \(i\) value of \(1\) because we'll consider both sucrose and NaCl to be non-ionizing solute (ideal behavior).
04

Set up the equations for mass fractions

Let \(x\) be the mass fraction of sucrose in the mixture, and then the mass fraction of NaCl will be \(1 - x\). We can create a system of equations to represent the mass of sucrose and NaCl in the solution. Sucrose: \(12x\) gram NaCl: \(20(1 - x)\) gram
05

Moles of Sucrose and NaCl

Calculate moles of sucrose and NaCl using the molar masses. Molar mass of sucrose: \(342.3 \frac{\mathrm{g}}{\mathrm{mol}}\) Molar mass of NaCl: \(58.44 \frac{\mathrm{g}}{\mathrm{mol}}\) Moles of sucrose: \(\frac{12x}{342.3}\) mol Moles of NaCl: \(\frac{20(1-x)}{58.44}\) mol
06

Calculate Total Molality

Calculate the total molality by using the equation: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \] Substitute the moles of sucrose and NaCl in the equation: \[ 0.229 = \frac{\frac{12x}{342.3} + \frac{20(1-x)}{58.44}}{1 \,\mathrm{kg}} \]
07

Solve for x (mass fraction of sucrose)

Solve the equation for x to find the mass fraction of sucrose in the original mixture: x = 0.422 Hence, the mass fraction of sucrose is \(42.2\%\) and that of NaCl is \((100 - 42.2) = 57.8\%\).
08

Calculate the mole fraction of sucrose

Now, using the previously calculated moles of sucrose and NaCl, we can calculate the mole fraction of sucrose: Moles of sucrose: \(\frac{12(0.422)}{342.3} = 0.0148\) mol Moles of NaCl: \(\frac{20(1-0.422)}{58.44} = 0.198\) mol Mole fraction of sucrose: \[ \chi_{sucrose} = \frac{\text{moles of sucrose}}{\text{moles of sucrose} + \text{moles of NaCl}} = \frac{0.0148}{0.0148 + 0.198} = 0.0697 \] Therefore, the mole fraction of sucrose in the original mixture is \(0.0697\).

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