Let f be the mass fraction of MgCl2 and (1-f) the fraction of NaCl. We can calculate the molarity of particles in the solution, C, by adding the effects of the two salts. MgCl2 makes 3 particles and the NaCl make 2. We assume that there is no Van t'Hoff factors beyond the number of particles (that is what is meant by ideality. Generally, salts are not ideal as concentrations get higher)
moles of particles from MgCl2: (.5 g * f ) ( 1mole/95.21 g) ( 3 particles/MgCl2) = .01574* f moles
moles of particles from NaCl (.5g *(1-f))* 1 mole/58.44 g)(2 particles/NaCl) = .01711*(1-f) moles
Concentration = ( .01574f + .01711(1-f))/1l = .01711 - .00137f mole/liter
The equation for the osmotic pressure into water from this solution is
π = CRT
C = π/RT
.001711 - .00137f = .430 atm/ ((.0821 liter atm/mole K)(298K)) = .01758
I'll leave the solving for f to you. Note that this is mass fraction, so 100f is the mass percent.
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