Manganese (II) chloride has a solubility of 5.75 M at 0 degrees C. If 125 mL of saturated MnCl2 is evaporated to dryness, what mass of MnCl2 will be left?

Question

Luna Chen · Accepted Answer

And so the solubility of #MnCl_2# is #5.75*mol*L^-1#..I am not sure if I believe that....I get a mass of approx. #90*g#...

We gots ................. #125*mLxx10^-3*L*mL^-1xx5.75*mol*L^-1=0.719*mol# ...the which represents a mass of .......#125.84*g*mol^-1xx0.719*mol=??*g# The question proposes quite an unrealistic scenario...

Noah Aldrich · Answer

undefined To find the mass of MnCl2 left when 125 mL of saturated MnCl2 is evaporated to dryness, first, calculate the number of moles of MnCl2 initially present using the given solubility.

Given:
Solubility of MnCl2 = 5.75 M
Volume of saturated solution = 125 mL = 0.125 L

Using the formula:
$ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution (in liters)}} $

Rearranging to solve for moles of solute:
$ \text{moles of solute} = \text{Molarity} \times \text{volume of solution (in liters)} $

Now, substitute the given values:
$ \text{moles of solute} = 5.75 \, \text{M} \times 0.125 \, \text{L} $

Calculate:
$ \text{moles of solute} = 0.71875 \, \text{moles} $

Since the compound is evaporated to dryness, the mass of MnCl2 left will be the mass of MnCl2 initially present. To find this mass, multiply the number of moles of MnCl2 by its molar mass.

The molar mass of MnCl2:
Molar mass of Mn = 54.938 g/mol
Molar mass of Cl = 35.453 g/mol (there are two Cl atoms in MnCl2)

Molar mass of MnCl2:
$ 54.938 \, \text{g/mol} + (2 \times 35.453 \, \text{g/mol}) = 125.844 \, \text{g/mol} $

Now, multiply the number of moles by the molar mass to find the mass of MnCl2 left:
$ \text{mass} = 0.71875 \, \text{moles} \times 125.844 \, \text{g/mol} $

Calculate:
$ \text{mass} = 90.38 \, \text{grams} $

So, when 125 mL of saturated MnCl2 is evaporated to dryness, 90.38 grams of MnCl2 will be left.