Calculate the boiling point of a solution prepared by adding 15 g of NaCl to 250 g of water(K_b for water + 0.512 K kg mol-1, Molar mass mas of NaCl = 58.44 g)

Moles of NaCl = 15/58.44

= 0.25 moles of NaCL

Kg of solvent = 250/1000 Kg

= 0.25 Kg of water

Therefore, Molality:

= 0.25/0.25

= 1 molal

For water change in BP:

ΔT_b = K_b * m

= 0.52 * 1

= 0.52 K

Since water boils at 373.15 K at 1.013 bar pressure, therefore the boiling of solution will be 373.15+0.52 = 373.67 K