Question

Solid mixture contains MgCl2 and NaCl. When 0.5000 g of this solid is dissolved in enough water to form 1.000 L of solution, the osmotic pressure at 25°C is observed to be 0.3930 atm. What is the mass percent of MgCl2 in the solid? (Assume ideal behavior for the solution.)

Answer 1

Answer:

83,40 (w/w) %

Explanation:

The osmotic pressure (π) is defined as:

π = iMRT

Where i is Van't Hoff factor (3 for MgCl₂ and 2 for NaCl), M is molarity, R is gas contant (0,082atmL/molK) and T is temperature (298,15K)

iM = π / RT

iM = 0,01607mol/L

It is possible to write:

3x+2y = 0,01607mol/L (1)

Where x are moles of MgCl₂ and y moles of NaCl.

-M = moles of each compund because M is molarity (moles/L) and there is 1,000L-

Knowing molar mass of MgCl₂ is 95,2 g/mol and for NaCl is 58,44g/mol:

x×95,211 + y×58,44 = 0,5000g (2)

Replacing (2) in (1):

3x+2(0,0086 - 1,629x) = 0,01607mol/L

-0,258x = -0,00113

x = 0,004380 moles of MgCl₂

In grams:

0,004380 moles of MgCl₂×(95,211g/mol) = 0,417g of MgCl₂

Mass percent is:

(0,4170g of MgCl₂/0,5000g of solid) ×100 = 83,40 (w/w) %

I hope it helps!