Question

The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So far this is what I have, but I cannot get the correct values for the mass of the solution to even begin to figure out the volume.

Answer 1

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of
`MgCl_2` (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of
`MgCl_2` (solute) = 3.37 g

First we have to calculate the mass of solute.

`\text{Mass of }MgCl_2=\text{Moles of }MgCl_2* \text{Molar mass of }MgCl_2`

`\text{Mass of }MgCl_2=3.37mole* 95.21g/mole=320.86g`

Now we have to calculate the mass of solution.

`\text{Mass of solution}=\text{Density of solution}* \text{Volume of solution}=1.25g/ml* 1000ml=1250g`

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

`Molality=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Mass of solvent}}=(3.37g* 1000)/(95.21g/mole* 929.14g)=0.0381mole/Kg`

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

`\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}* 100=(320.86)/(1250)* 100=25.67\%`

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

`\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}* 100=(320.86)/(1000)* 100=32.086\%`

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.