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?

Feb 26, 2017

$\text{At least.........} 5.15 \cdot g$

Explanation:

Originally, you have a $35 \cdot m L$ volume of $0.500 \cdot m o l \cdot {L}^{-} 1$ ${K}_{2} C {r}_{2} {O}_{7}$.

This represents a molar quantity of $35 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.500 \cdot m o l \cdot {L}^{-} 1 = 0.0175 \cdot m o l$ with respect to $\text{potassium dichromate.}$

This molar quantity represents a mass of $0.0175 \cdot m o l \times 294.18 \cdot g \cdot m o {l}^{-} 1 = 5.15 \cdot g$.

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