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

Question

2745 views around the world

You can reuse this answer
Creative Commons License

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*mL# volume of #0.050*mol*L^-1# concentration with respect to dichromate?