How much K2Cr2O7(mol.wt.=294.19) is required to prepare 1L of 0.1 M solution in acid medium

Asked by Anil | 17th May, 2017, 02:28: PM

Expert Answer:

Equivalent of potassium dichromate

= Molar mass / 6

= 294.19/6

=49.30

For  0.1 N = 1/10 N

Weight of K2Cr2O7 required to prepare one litre 1/10 N

= Equivalent of potassium dichromate / 10

= 49.30 / 10

=4.93 g

Answered by Prachi Sawant | 17th May, 2017, 03:17: PM