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.

Boiling point of solution is

Explanation:

Cane sugar is a non-volatile solute.

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

Where, is elivation in boiling point of solution, 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,

610 g of cane sugar = moles of cane sugar

                                  = 1.78 moles of cane sugar

So, molality of solution (m) =

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

So, boiling point of solution =              

it means there are 3 atoms of oxygen