Respuesta :
Answer:
Boiling point of solution is [tex]100.65^{0}\textrm{C}[/tex]
Explanation:
Cane sugar is a non-volatile solute.
According to Raoult's law for a non-volatile solute dissolved in a solution-
[tex]\Delta T_{b}=K_{b}.m[/tex]
Where, [tex]\Delta T_{b}[/tex] is elivation in boiling point of solution, [tex]K_{b}[/tex] 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, [tex]K_{b}=0.51^{0}\textrm{C}.kg.mol^{-1}[/tex]
610 g of cane sugar = [tex]\frac{610}{342.3}[/tex] moles of cane sugar
= 1.78 moles of cane sugar
So, molality of solution (m) = [tex]\frac{1.78}{1.4}mol.kg^{-1}=1.27mol.kg^{-1}[/tex]
Plug in all the values in the above equation, we get-
[tex]\Delta T_{b}=0.51^{0}\textrm{C}.kg.mol^{-1}\times 1.27mol.kg^{-1}=0.65^{0}\textrm{C}[/tex]
So, boiling point of solution = [tex](100+0.65)^{0}\textrm{C}=100.65^{0}\textrm{C}[/tex]
