When 0.6 g of urea is dissolved in 100 g water then what is the boiling point of water? (Kb for water = 0.52 Km-1 and normal boiling point of water = 100oC)
Answer & explanation
Correct answer: option 3
Elevation in boiling point, ΔTb = \(\frac{K_b×1000×W_2}{W_1 × M_2}\)
= \(\frac{0.52×1000×0.6}{100×60}\) = 0.052
Normal boiling point of water = 100oC = 373 K
Boiling point of of 0.6 g urea in 100 g of water = 373+0.052 = 373.052 K