What mass of sodium dichromate is required to prepare a 250*mL volume of 0.050*mol*L^-1 concentration with respect to dichromate?

1 Answer
May 1, 2016

Approx. $3 \cdot g$ of the anhydrous salt.

Explanation:

$\text{Moles of sodium dichromate required}$ $=$ $250 \times {10}^{-} 3 L \times 0.05 \cdot m o l \cdot {L}^{-} 1$ $=$ $0.0125 \cdot m o l$

To get the required mass, we simply mulitply this number by the molar mass of the anhydrous salt:

$\text{Mass}$ $=$ $0.0125 \cdot m o l \times 261.97 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g