If you wish to prepare a 1.13 M solution of NaNo3, to what volume (in liters) would you have to dilute 25.4 mL of 3.21 M NaNO3?

Answer is 72.15 or 0.07215

Explanation:

Given,

Initial molarity M1 = 3.21M

Initial volume V1 = 25.4mL

Final molarity of M2 = 1.13 M

Final volume V2 = ?

M1×V1=M2×V2

M1×V1/M2=V2

3.21M×25.4/1.13=V2

V2=72.15 mL. 1 mL=0.001 L

V2= 0.07215

the answer would be a

i think that the answer is c