Question

A stock solution of MgCrO₄ is available to prepare solutions that are more dilute. Calculate the volume, in mL, of a 2.0-M solution of MgCrO₄ required to prepare exactly 250mL of a 0.510-M solution of MgCrO₄.

_____ mL?

A solution is made by dissolving 19.6 g of iron(II) bromide, FeBr₂, in enough water to make exactly 500 mL of solution. Calculate the concentration (molarity) of FeBr₂ in mol/L (M).

_____ M FeBr₂

Calculate the mass, in grams, of CoBr₂ required to prepare exactly 100 mL of a 0.510-M solution of CoBr₂.

_____g

Answer 1

When we dilute a solution, that is increase the volume of solution without increasing the mols of solute, then the concentration of the diuted solution is less than the original one.

Now if

the initial concentration of the solution taken = M1

the initial volume of the solution taken = V1

the final volume after dilution becomes = V2

And the final concentration after dilution becomes = M2

then the relation between them is-

M1V1 = M2V2

So based on this concept-

1-

Given for the MgCrO4 solution-

initial concentration (M1) = 2.0-M

initial volume (V1) = ?

Final concentration (M2) = 0.510-M

Final volume (V2) = 250mL

Then putting the values-

M1V1 = M2V2

V1 = M2V2/ M1

= 0.510-M * 250mL/ 2.0-M

= 63.75‬ mL

That means the initial volume of MgCrO4 taken = 63.75‬ mL

2-

Given mass of FeBr2 taken = 19.6 g

So mols of FeBr2 taken = mass / molar mass of FeBr2

= 19.6 g / 215.65 g/mol

= 0.091 mols

Now total volume of FeBr2 solution = 500 mL

So Molarity = mols of solute / volumee of solution in L

= 0.091 mols /500 mL

= 0.091 mols / 0.5 L

= 0.182‬ mol/L

= 0.182‬ M

3-

Given concentration of CoBr2 solution = 0.510-M

volume of CoBr2 solution = 100 mL

So mols of CoBr2 present = concentration * volume

= 0.510-M * 100 mL

=  0.510 mols/ 1000 mL * 100 mL

= 0.051‬ mols

So mass of CoBr2 present = moles * molar mas

= 0.051‬ mols * 218.74 g/mol

= 11.16 g