Question

What is the melting point of a 3L aqueous solution that contains 100g of MgCl2? kf H2O=1.86 rhoH2O=1gmL

Answer 1

Answer:

Melting point of aqueous solution = -10.32 °C

Explanation:

$\Delta T_f=i \times k_f \times m$

Where,

ΔT_f = Depression in freezing point

k_f = molal depression constant

m = molality

Formula for the calculation of molality is as follows:

$m=\frac{Mass\ of\ solute\ (kg)}{molecular\ mass\ of\ solute \times mass\ of\ solvent}$

density of water = 1 g/mL

density = mass/volume

Therefore,

mass = density × volume

volume = 3 L = 3000 mL

Mass of water = 1 g/mL × 3000 mL

                        = 3000 g

$Molality(m)=\frac{100\times1000}{18\times 3000} \\=1.85\ m$

van't Hoff factor (i) for MgCl2 = 3

Substitute the values in the equation (1) to calculate depression in freezing point as follows:

$\Delta T_f=i \times k_f \times m\\=3\times 1.86 \times 1.85\\=10.32\ °C$

Melting point of aqueous solution = 0 °C - 10.32 °C

                                                          = -10.32 °C

Answer 2

Answer:

The melting point of the solution is - 1.953 °C

Explanation:

In an ideal solution, the freezing point depression is computed as follows:

$ΔT_f = k_f \times b \times i$

where:

$ΔT_f$ is the freezing-point depression

$k_f$ is the cryoscopic constant, in this case is equal to 1.86

b is the molality of the solution

i is the van't Hoff factor, number of ion particles per individual molecule of solute, in this case is equal to 3

Molality is defined as follows:

b = moles of solute/kg of solvent

Moles of solute is calculated as follows:

moles of solute = mass of solute/molecular weight of solute

In this case there are 100 g of solute and its molecular weight is 35.5*2 + 24 = 95 g/mole. So, the moles are:

moles of solute = 100 g/(95 g/mol) = 1.05 moles

The mass of solvent is computed as follows:

mass of solvent = density of solvent * Volume of solvent

Replacing with the data of the problem we get:

mass of solvent = 1 kg/L*3 L = 3 kg

Finally, the molality of the solution is:

b = 1.05/3 = 0.35 mol/kg

Then, the freezing-point depression is:

$ΔT_f = 1.86 \times 0.35 \times 3$

$ΔT_f = 1.953 C$

The freezing-point depression is the difference between the melting point of the pure solvent (here water) and the melting point of the solution. We know that the the melting point of water is 0 °C, then:

melting point of water - melting point of the solution = 1.953 °C

melting point of the solution  = 0 °C - 1.953 °C = - 1.953 °C