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anonymous

  • one year ago

**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.

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  1. anonymous
    • one year ago
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    @dan815 @Loser66 @Luigi0210 @BTaylor

  2. aaronq
    • one year ago
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    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

  3. anonymous
    • one year ago
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    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.

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