**Please just don't make me go through a long process. I just need to finish the class.**
A solution is made by dissolving 3.8 moles of sodium chloride (NaCl) in 185 grams of water. If the molal boiling point constant for water (Kb) is 0.51 °C/m, what would be the boiling point of this solution? Show all of the work needed to solve this problem.

Question

**Please just don't make me go through a long process. I just need to finish the class.** 
A solution is made by dissolving 3.8 moles of sodium chloride (NaCl) in 185 grams of water. If the molal boiling point constant for water (Kb) is 0.51 °C/m, what would be the boiling point of this solution? Show all of the work needed to solve this problem.

anonymous · Answer

@dan815 @Loser66 @Luigi0210 @BTaylor

aaronq · Answer

Find the molality then use: $\Delta T=i*m*K_b$

Where $\Delta T$ is the $\sf \color{red}{change}$ in the temperature from the pure solvent's boiling point

anonymous · Answer

molarity of NaCl solution = 3.8 mol / 0.185 kg = 20.54 molal. Since NaCl is ionic, the concentration of ions in the solution is 2X10.54 = 41.1 molal 

DTb = Kb m = 0.51C/m X 41.1 = 21 C 

So, the boiling point of the solution will be 100C + 21 C = 121 C.