In a titration, 14.1 ml of 0.0500 m na2s2o3 was required to turn 15.0 ml of an aqueous solution containing i2, i3-, and starch from dark blue to colorless.
how many moles of i2 and i3- were present in the solution altogether?

The correct formula to use for this problem is
VACA= VBCB, where VA= volume of the acid, CA= concentration of acid, VB =volume of base and CB = concentration of base.
15*CA = 14.1*0.05 =
CA = 14.1*0.05/15 = 0.047
Therefore, the number of moles (concentration) present in the solution is approximately 0.05M.

Given:

Concentration of titrant: 0.0500 M Na2S2O3
Volume of titrant: 14.1 mL
Volume of aqueous solution: 15.0 mL

Solve for the concentration of the aqueous solution using this formula:

C(sol'n) * V (sol'n) = CNa2S2O3 * VNa2S2O3

C(sol'n) = CNa2S2O3 * VNa2S2O3 / V(sol'n)

C(sol'n) = 0.05M * 14.1 mL / 15 mL
             = 0.047 M

Number or moles = 0.047 mol/L * 15 ml/1000ml/L
                              = 7.05x10^-4 moles 

Ch4 is nonpolar and insoluble in water. since all of the hydrogen atoms are evenly spread around the carbon atom even through the c h bond is polar all the poles cancel each other out.