Please be sure to explain your answer One solution is formed

Please be sure to explain your answer

One solution is formed by mixing 12.5 g of sucrose (C12H2201) into 100 g of water. A second solution is formed by mixing 11.1 g of ferric chloride (FeCl;) into 100 g of water. Which of these solutions will have a lower boiling point? Show your work and explain your answer. (4 pts)

Solution

Solution 1 has 12.5 g of sucrose (C₁₂H₂₂O₁₁) in 100 g water.

Molar mass of sucrose = (12*12.01 + 22*1.008 + 11*15.9994) g/mol = 342.2894 g/mol.

Molality of solution 1 = (moles of sucrose)/(kg of water) = (mass of sucrose/molar mass of sucrose)/(kg of solvent) = [(12.5 g)/(342.2894 g/mol)]/[(100 g)*(1 kg/1000 g)] = 0.3652 m.

Solution 2 has 11.1 g of ferric chloride (FeCl₃) in 100 g water.

Molar mass of ferric chloride = (1*55.845 + 3*35.453) g/mol = 162.204 g/mol.

Molality of solution 2 = (moles of ferric chloride)/(kg of water) = (mass of ferric chloride/molar mass of ferric chloride)/(kg of solvent) = [(11.1 g)/(162.204 g/mol)]/[(100 g)*(1 kg/1000 g)] = 0.6843 m.

The boiling point elevation of a solution is given as T_b = T_s – T_p where T_p = 100°C is the boiling point of pure water and T_s is the boiling point of the solution.

Again, the boiling point elevation of a solution is related to the molality of the solution as

T_b = k_b*(molality of solution) where k_b = boiling point elevation constant is constant for a particular solvent.

With the above two expressions in hand, we can write,

T_b1 = k_b*(0.3652 m)

T_b2 = k_b*(0.6843 m)

where the subscripts 1 and 2 denote solutions 1 and 2 respectively.

Since solution 1 has a lower molality as compared to solution 2, solution 1 will have the lower boiling point elevation and hence, the lower boiling point.