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

1. User Answers 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
∴T20 =  2 * 0.26 *1.86
∴T2 = 0.967 °C 
2. User Answers 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