What is the melting point of a solution in which 3.5 grams of sodium chloride is added to 230 mL of water?

• superman1987: we are going to use this equation:
  
  ΔT = - i m Kf
  
  when m is the molality of a solution 
  
  i = 2
  
  and ΔT is the change in melting point = T2- 0 °C
  
  and Kf is cryoscopic constant = 1.86C/m
  
  now we need to calculate the molality so we have to get the moles of NaCl first:
  
  moles of NaCl = mass / molar mass
  
                           = 3.5 g / 58.44 
  
                           = 0.0599 moles
  
  
  when the density of water = 1 g / mL and the volume =230 L
  
  ∴ the mass of water = 1 g * 230 mL = 230 g = 0.23Kg 
  
  now we can get the molality = moles NaCl / Kg water
  
                                                  =0.0599moles/0.23Kg
  
                                                  = 0.26 m
  
  ∴T2-0 =  - 2 * 0.26 *1.86
  
  ∴T2 = -0.967 °C
• jbiain:

  Answer:

  0.952 °C

  Explanation:

  The change in melting point is computed as:

  ΔT = k*m*i

  where ΔT is the difference between the melting point of water and of solution, k is a constant (1.86 °C*kg/mol for water), i is the van't Hoff factor (equal to 2 for sodium chloride because 2 ions are obtained after its dissolution), and m is the molality of the solution.

  Molar mass of sodium chloride: 58.44 g/mol

  Moles of of sodium chloride: mass / molar mass 3.5/58.44 = 0.059 mol

  Density of water 1 kg/L

  230 mL of water are equivalent to 0.23 L

  mass of water: density * volume = 1*0.23 = 0.23 kg

  Molality of the solution: m = moles of solute/ kg of solvent = 0.059/0.23 = 0.256

  Finally:

  ΔT = 1.86*0.256*2 = 0.952 °C

  Water melting point: 0 °C

  So, the solution melting point is: 0 - 0.952 = 0.952 °C