Answer to Question #223920 in Chemistry for Lebo

Question

A CaCO3-MgCO3 solid solution contains 3 weight percentage Mg. What is the mole fraction of MgCO3 in the solid?

Answer 1

Solution:

The molar mass of CaCO₃-MgCO₃ is 184 g/mol.

Suppose that 184 g (or 1 mol) of a CaCO₃-MgCO₃ solid solution are given.

Therefore,

m(Mg) = [w(Mg) × m(CaCO₃-MgCO₃ solid solution)] / 100% = (3% × 184 g) / 100% = 5.52 g Mg

The molar mass of Mg is 24 g/mol.

Therefore,

(5.52 g Mg) × (1 mol Mg / 24 g Mg) = 0.23 mol Mg

MgCO₃ → Mg

According to the equation above: n(Mg) = n(MgCO₃) = 0.23 mol

Thus,

χ(MgCO₃) = n(MgCO₃) / n(CaCO₃-MgCO₃)

χ(MgCO₃) = (0.23 mol) / (1 mol) = 0.23

χ(MgCO₃) = 0.23

Answer: The mole fraction of MgCO₃ in the solid is 0.23