Question

A solid mixture contains MgCl₂ (molar mass = 95.218g) and NaCl (molar mass= 58.443g). When 0.5000g of this solid is dissolved in enough water to form 1.000L of the solution, the osmotic pressure at 25C is observed to be 0.3950atm. What is the mass percent of MgCl₂ in a solid? Assume ideal behavior for a solution.

Short answer

Mass percent of MgCl₂is 72%.

Step-by-step solution

Step 1- Calculating Apparent molarity:

Osmotic pressure = molarity x R x T

0.3950 atm = (x) X 0.08206 X 298.15

x = 0.016M

–Balancing the charges:

M^-_Cl = 2M_Mg2++ M_Na+

– Calculating mass percent of MgCl2:

Total ion concentration must be equal to the concentration of all ion

0.016 mol/L = M_Mg⁺²+ M_Na⁺ + M_Cl^-

0.016 = 3Mg⁺²+ 2M_Na⁺

Let the mass of MgCl₂ is x and y be the mass of NaCl, then we calculate molarity in terms of x and y

x + y = 0.5000g

M_Mg⁺²= x/95.218

M_Na⁺= y/58.443

Adding these equations and equating it to total ion concentration

3x/95.218 + 2y/58.443 = 0.016

= 3x + (3.25) y = 1.537

Solving equation, we get y= 0.14 and x= 0.36 g Thus, mass of NaCl is 0.14 g and MgCl₂ is 0.36 g

So, mass percent of MgCl₂ is

0.36g/0.5000g x 100%

= 72%