Question

Density of a solution?

A solution of magnesium hydroxide is 67.0% by mass. If it is 8.8 mol/L, what is the density of the solution?

Answer 1

See the process steps below;

Explanation:

Recall;

`"Density" = "mass"/"volume" rArr d = m/v`

`"Magnesium Hydroxide" = Mg(OH)_2`

`"Molar mass" = 58 gmol^-1`

`"Composition" = "mass"/"molar mass" xx 100`

`rArr 67% = m/58`

`rArr 67/100 = m/58`

`rArr 100m = 67 xx 58`

`rArr 100m = 3886`

`rArr m = 3886/100`

`rArr m = 38.86g`

Since we are given the concentration which is; `8.8molL^-1`

`:. c = 8.8molL^-1`

`"no of moles" = "mass"/"molar mass" rArr n = m/(mM)`

`rArr n = (38.86g)/(58gmol^-1) = 0.67mols`

`:. n = 0.67mols`

Recall;

`c = n/v`

`v = n/c`

`v = (0.67mols)/(8.8molsL^-1)`

`v = (0.67cancel(mols))/(8.8cancel(mols)L^-1)`

`v = 0.076L`

Hence;

`d = m/v`

`d = (38.86g)/(0.076L)`

`d = 511.316gL^-1`

Hope this helps!

Answer 2

If this were possible, the density of the solution would be 0.815 g/mL.

Explanation:

A correct answer is impossible because the solubility of `"Mg(OH)"_2` is only 6 mg/L.

However, here is how I would calculate the density of the solution.

Step 1. Calculate the mass of `"Mg(OH)"_2` in 1 L of 8.8 mol/L solution

`"Mass of Mg(OH)"_2 = 1 color(red)(cancel(color(black)("L solution"))) × (8.8 color(red)(cancel(color(black)("mol Mg(OH)"_2))))/(1 color(red)(cancel(color(black)("L solution")))) × ("58.32 g Mg(OH)"_2)/(1 color(red)(cancel(color(black)("mol Mg(OH)"_2)))) = "513 g Mg(OH)"_2`

Step2. Calculate the mass of a solution containing 513 g `"Mg(OH)"_2`

`"Mass of solution" = 513 color(red)(cancel(color(black)("g Mg(OH)"_2))) × "100 g solution"/(63.0 color(red)(cancel(color(black)("g Mg(OH)"_2)))) = "810 g solution"`

Step 3. Calculate the density of the solution

You have 513 g `"Mg(OH)"_2` in a solution with a mass of 815 g and a volume of 1 L.

`ρ = "mass"/"volume" = "810 g"/"1000 mL" = "0.81 g/mL"`

Even this answer does not make sense. Surely a solution this concentrated would have a density greater than 1 g/mL.