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

Boiling point of solution is
`100.65^(0)\textrm{C}`

Step-by-step 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 =
`(610)/(342.3)` moles of cane sugar

= 1.78 moles of cane sugar

So, molality of solution (m) =
`(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)* 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}`