Question

the amount of K2cr2o7 (eq.wt.49.04) required to prepare 100ml of its 0.05 N solution is​

Answer 1

Explanation:

Normality =

vol. of solution in L

No. of gm-eq of solute

we have,

gm-eq wt.=49.04,volum = 0.1L and N=0.05

now,

0.05=

0.1

No. of gm-eq of solute

No. of gm-eq of solute=0.005

0.005=

eq. wt

wt. of solute

wt. of solute=0.005×49.04

=0.2452 g