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

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Answer 1 · 2016-05-01T14:55:53

Approx. $3*g$ of the anhydrous salt.

$"Moles of sodium dichromate required"$ $=$ $250xx10^-3Lxx0.05*mol*L^-1$ $=$ $0.0125*mol$

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

$"Mass"$ $=$ $0.0125*molxx261.97*g*mol^-1$ $=$ $??g$

What mass of sodium dichromate is required to prepare a 250*mL250*mL volume of 0.050*mol*L^-10.050*mol*L^-1 concentration with respect to dichromate?