Question

A solution is made by dissolving 1.00 moles of sodium chloride (NaCl) in 155 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.

Answer 1

Answer:

106.6 °C  

Step-by-step explanation:

The formula for boiling point elevation ΔTb is

ΔTb = iKb·b

where

i     = the van't Hoff i# factor

Kb = the molal boiling point elevation constant

b   = the molal concentration of the solution

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Data

  i = 2, because 1 mol of NaCl gives 2 mol of ions in solution.

Kb = 0.51 °C·mol·kg⁻¹

 b = 1.00/0.155

 b = 6.452 mol·kg⁻¹

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Calculations

ΔTb = 2 × 0.51 × 6.452

ΔTb = 6.58°C

 Tb = Tb° + ΔTb

 Tb = 100 + 6.58

 Tb = 106.6 °C