Because of the lack of reactives in a laboratory, you wish to prepare a solution of agno3 0.1 n, given two solutions, the first 0.0825n and the second 0.7280n. what volume of each of them you'll have to mix to obtain 1lt of the solution you want?

Given:
c1=0.0825 M
c2=0.7280 M
c3=0.1 M
V3 =1 dm3
Setting up the equation:
c1V1+c2V2 = c3V3
now you work a little magic:
V1+V2=V3
Solving for V1
V1= V3 - V2
substitute V1 in first equation:
c1(V3-V2) + c2V2=c3V3

Solving: 

c1V3-c1V2 + c2V2=c3V3

Rearranging the values:

c1V3-c3V3=c1V2-c2V2

V3(c1-c3)=V2(c1-c2)

So,
V2 = V3(c1-c3) / c1-c2
Substituting known values:
V2= 1 dm3 * (0.0825-0.1) / 0.0825 - 0.7280
and you get:
V2= 0.027 dm3
Hence,
V1 = V3-V2 V1
     = 0.973 dm2