Question

You take 20.0 g of sucrose (${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$) and NaCl mixture and dissolve it in 1.00 L of water. The freezing point of this solution is found to be_{$-0.426°\text{C}$}. Assuming ideal behavior, calculate the mass percent composition of the original mixture and the mole fraction of sucrose in the original mixture.

Short answer

The mass percent composition of sucrose and sodium chloride is 2.9% and 20.0% respectively. The mole fraction of sucrose in the original mixture is 0.12.

Step-by-step solution

Molality of the solution

According to the question, it is given that

Mass of sucrose and NaCl mixture = 20.0 g

Mass of water = 1.00 kg

The freezing point of the solution = _{$-0.426°\text{C}$}

_{${\text{k}}_{\text{f}}\text{for water = 1.86}°\text{C/m}$}

The freezing temperature of pure water =$0°\text{C}$

Thus, the freezing point of the solution

= _{${\text{T}}_{\text{Pure}}{\text{-T}}_{\text{Solution}}{\text{=K}}_{\text{f}}\times \text{m}$}

= _{$\text{0 - ( - 0.426)=1.86}\times \text{m}$}

m = _{$\frac{\text{0.426}}{1.86}\text{= 0.226 mol/kg}$}

Moles of the solution

The number of moles of the solution will be _{$0.226\text{mol/kg}\times \text{1kg = 0.226 mol}$}

Let the number of moles of sucrose is and the number of moles be y, then

_{$\text{x + y = 0.229}$}

Or _{$\text{x = 0.229 - y}$} …. (1)

Now, we know that the molar mass of sucrose is _{$\text{12}\times \text{12 + 22}\times 1+11\times \text{16 = 342 g/mol}$}

And the molar mass of NaCl is _{$\text{23}\times \text{1 + 37.5}\times 1\text{= 58.5 g/mol}$}

So, the total mass of the solution can be written as:

_{$\text{x}\times \text{342 + y}\times \text{58.4 = 20}$}

$\text{342x + 58.4y = 20}$ …. (2)

By putting the value of x from equation (1), we will get:

_{$\text{342(0.229 - y) + 58.4y = 20}$}

_{$\text{78.3 - 342y + 58.4y = 20}$}

_{$\text{283.6y = 58.3}$}

y = 0.20

Thus, x = 0.229 – 0.20 = 0.029

Total number of moles will be 0.20 + 0.029 = 0.229

Mole percent of the solution

The mole percent of sucrose = $0.029\times \text{100}$ = 2.9%

The mole percent of NaCl = $0.20\times \text{100}$ = 20.0%

Mole fraction of sucrose

The mole fraction of sucrose = $\frac{\text{Moles of sucrose}}{\text{Total number of moles}}$ = _{$\frac{0.029}{0.229}\text{= 0.12}$}