When 0.6 g of urea is dissolved in 100 g water then what is the boiling point of water? (K_b for water = 0.52 Km^-1 and normal boiling point of water = 100^oC) 

373.052 K

Elevation in boiling point, ΔT_b = \(\frac{K_b×1000×W_2}{W_1 × M_2}\)

                                            = \(\frac{0.52×1000×0.6}{100×60}\) = 0.052

Normal boiling point of water = 100^oC = 373 K

Boiling point of of 0.6 g urea in 100 g of water = 373+0.052 = 373.052 K