Question

If 35 ml of the original 0.500 M `K_2Cr__2O_7(aq)` solution was evaporated to dryness, what mass `KaCr_2O_7(s)` would remain in the evaporating dish?

Answer 1

`"At least........."5.15*g`

Explanation:

Originally, you have a `35*mL` volume of `0.500*mol*L^-1` `K_2Cr_2O_7`.

This represents a molar quantity of `35*mLxx10^-3*L*mL^-1xx0.500*mol*L^-1=0.0175*mol` with respect to `"potassium dichromate."`

This molar quantity represents a mass of `0.0175*molxx294.18*g*mol^-1=5.15*g`.

Of course, the salt might be solvated by several water molecules. This would increase the mass.