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.

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: \[m = \frac{ΔT_f}{K_f} = \frac{0.426}{1.86} = 0.2289\, mol/kg\]

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.

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: \[0.02368 + y = 0.2289\] \[y = 0.2052 \, mol/kg\]

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.