Question

What is the boiling point of a solution produced by adding 610 g of cane sugar (molar mass 342.3 g/mol) to 1.4 kg of water? For each mole of nonvolatile solute, the boiling point of 1 kg of water is raised 0.51 ∘C.

Answer 1

Answer:

Boiling point of solution is $100.65^{0}\textrm{C}$

Explanation:

Cane sugar is a non-volatile solute.

According to Raoult's law for a non-volatile solute dissolved in a solution-

                              $\Delta T_{b}=K_{b}.m$

Where, $\Delta T_{b}$ is elivation in boiling point of solution, $K_{b}$ is ebbulioscopic constant of solvent (how much temperature is raised for dissolution of 1 mol of non-volatile solute) and m is molality of solution.

Here, $K_{b}=0.51^{0}\textrm{C}.kg.mol^{-1}$

610 g of cane sugar = $\frac{610}{342.3}$ moles of cane sugar

                                  = 1.78 moles of cane sugar

So, molality of solution (m) = $\frac{1.78}{1.4}mol.kg^{-1}=1.27mol.kg^{-1}$

Plug in all the values in the above equation, we get-

$\Delta T_{b}=0.51^{0}\textrm{C}.kg.mol^{-1}\times 1.27mol.kg^{-1}=0.65^{0}\textrm{C}$

So, boiling point of solution = $(100+0.65)^{0}\textrm{C}=100.65^{0}\textrm{C}$