% of NaCl in this mixture (NaCl is dissociated) -I've already done an equation?

Question

1) We have a mixture of: NaCl(s) and C12H22O11(s).
2) 15.0g of this mixture is dissolved in enough water to form 500mL of solution.
3) The osmotic pressure of this solution is 6.41atm and 25°C

With this information, I need to know the weight % of NaCl.

I did this (see below) and my teacher told me that I needed to consider that the NaCl dissociates in water. What should I fix? Thank you.

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Osmotic Pressure=Pi M=Molarity R=Constant T=Temperature
Pi=MRT
M=Pi/RT
M=6.41atm/(0.082)(298.15K) = 0.262 M

0.262 M= 0.262 moles mixture / 1 L of solution
We have .5 L of solution, so we have 0.131 moles mixutre

Equation
x g NaCl * (mol NaCl/58.5g NaCl) + (15-x)g C12H22O11 *(mol C12H22O11/342g) = 0.131 mol

NaCl=x=6.15g
C12H22O11=15-x=8.85g

But I didn't consider that NaCl is dissociated.
Thanks for your answer.

Answer 1

You need to consider that NaCl dissociates into 1 Na+ and 1 Cl-, thus the number of mole particles coming from NaCl is going to be double than what you wrote.
So you would have 2x/58.5+15-x/342 =0.131