Question

A student determines that a 9.8 g mixture of MgCl₂ (s) and NaNO3(s) contains 0.050 mol of

Answer 1

mass MgCl₂ = mol x MM MgCl₂ = 0.05 x 95.211 g/mol = 4.76 g

mass Cl in MgCl₂ :

= (2 x AM Cl)/MM MgCl₂ x mass MgCl₂

= (2 x 35.5 g/mol)/95.211 g/mol x 4.76

= 3.55 g

% mass Cl in the mixture :

= (mass Cl / mass mixture) x 100%

=  3.55 / 9.8 x 100%

= 36.22%