Question

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

Answer 1

0.952 °C

Step-by-step 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

Answer 2

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