Question

What is the boiling point of a solution of 115.0 g of sucrose, \({\bf{C_{12}H_{22}O_{11} }}\), in 350.0 g of water?

1. Outline the steps necessary to answer the question
2. Answer the question

Short answer

The boiling point of a solution of 115.0 g of sucrose, \({\bf{C_{12}H_{22}O_{11} }}\), in 350.0 g of water is 100.5 degrees Celsius.

Step-by-step solution

Boiling Point

Thetemperature at which liquid vapor pressure equals atmospheric pressure is referred to as the boiling point.

Subpart (a)

The below mentioned steps are followed:

i) Calculate the molality of sucrose,

\({\bf{Molality = }}\frac{{{\bf{Number}}\,{\bf{of}}\,{\bf{moles}}\,of\,solute\,sucrose}}{{{\bf{Mass}}\,{\bf{of}}\,{\bf{Solvent(kg)}}}}\)

ii) Equation to determine the boiling point is:

\({\bf{\Delta Tb = i \times (Kb) \times m}}\),

Where,

m = molality,

\({\bf{Kb}}\)is the molal elevation constant of the solvent

\({K_b} = 0.52^\circ C/m\)

i - the Van't Hoff factor = 1 for sucrose.

Subpart (b)

i)Molality of the solution:

\(\begin{align}m &= \frac{{115.0\,g}}{{342.3\,g/mol}} \times \frac{1}{{0.350\,kg}}\\ &= 0.959m\end{align}\)

ii) The boiling point of pure water is 10 degrees Celsius.

\(\begin{align}\Delta Tb &= i \times (Kb) \times m\\{T_b} - 100^\circ C &= 1 \times 0.52^\circ C/m \times 0.959m\\{T_b} &= 100.49^\circ C\\ &= 100.5^\circ C\end{align}\)