Question

the boiling point of water is 100 degree C.Calculate the boiling point of aquous solution containing 0.6g of urea in 100g of water (kb for water).0.52kkg per mol

Answer 1

Answer ; from elevation in boiling,

where  is elevation in boiling point,  is ebullioscopic constant and m is molality of solute.

mass of urea = 0.6 g

molar mass of urea (solute) = 60g/mol

so, number of mole of urea = 0.6/60 = 0.01

mass of solvent (water) = 100g

so, molality of urea = number of mole of urea/mass of solvent in kg

= 0.01 × 1000/100 = 0.1 molal

given, ebullioscopic constant,  = 0.52K/molal or 0.52°C/molal

then, ∆Tb = 0.52 × 0.1 = 0.052K

so, boiling point of solution, Tb = T + 0.052°C

= 100°C + 0.052 °C = 100.052°C