Question

the elevation in boiling point of aq solution of cane sugar when 34.2g of cane sugar dissolved in 1kg of water​

Answer 1

you didn't mention about ebullioscopic constant of cane sugar solution. Let's consider that Kb is ebullioscopic constant.

now find molality of solution.

given, mass of can sugar = 34.2g

molar mass of sugar = 342g/mol

so, number of mole = 34.2/342 = 0.1mol

mass of solvent (water) = 1kg

now, molality = number of mole of solute/mass of solvent in kg

= 0.1mol/1kg = 0.1m

now using formula

elevation in boiling point,∆Tb = Kb × m

= Kb × 0.1

= (0.1Kb) Kelvin

now you can easily get elevation of boiling point by using Kb value.