Calculate Mass % of MgCl2

Question

A solid mixture contains MgCl2 and NaCl. When 0.5000g of this solid is dissolved in enough water to form 1.000L of solution, the osmotic pressure at 25 degrees Celsius is observed to be 0.4030 atm. What is the mass percent of MgCl2 in the solid? (Assume ideal behavior for the solution.)

Answer 1

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.

Please consider a tutor. Take care.