x is not in mL..it is just a pure number. Quantity[x, "ml"]
is the 'thing' that is in mL.
To get what you want, you need to recast your Solve command as
Solve[x Quantity[5, "mol"] + (Quantity[250, "ml"] - x) Quantity[7,
"mol"] == Quantity[250, "ml"] Quantity[6, "mol"], x]
However, the result is a bit strange looking.
{{x -> Quantity[1/8000, ("Meters")^3]}}
It is, however, correct. Let's convert it to numerical form
Solve[
x Quantity[5, "mol"] + (Quantity[250, "ml"] - x) Quantity[7,
"mol"] == Quantity[250, "ml"] Quantity[6, "mol"], x] // N
gives
{{x -> Quantity[0.000125, ("Meters")^3]}}
Which looks right since 1 mL= 0.000001 m^3
How to force Mathematica to stick to mL, however, is another question.
Of course you could get back to mL by using UnitConvert
. For example
UnitConvert[
x /. Solve[
x Quantity[5, "mol"] + (Quantity[250, "ml"] - x) Quantity[7,
"mol"] == Quantity[250, "ml"] Quantity[6, "mol"],
x] [[1]], "Milliliters"]
gives
Quantity[125, "Milliliters"]