What is the melting point of a 3l aqueous solution that contains 100g of mgcl2? kf h2o=1.86 rhoh2o=1gml

The melting point of the solution is - 1.953 °C

Explanation:

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

where:

is the freezing-point depression

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:

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