Answer:
8.909*10^-4 moles
Explanation:
The mixture contains MgCl[tex]_{2}[/tex] and NaCl with a total mass of 0.4015 g. The mass of precipitate AgCl(aq) is 1.032 g and it has a molar mass of 143.32 g/mol. Therefore, the moles of the precipitate is:
n = 1.032/143.32 = 7.2007*10^-3 moles
Molar mass of NaCl = 58.44 g/mol and the molar mass of MgCl[tex]_{2}[/tex] is 95.21 g/mol. Let the mass (g) of MgCl[tex]_{2}[/tex] in the original mixture be 'x'. Thus:
7.2007*10^-3 = (0.4015-x)/58.44 + 2x/95.21
(7.2007*10^-3)*58.44*95.21 = 95.21(0.4015-x) + 2x(58.44)
40.06505 = 38.227 -95.21x + 116.88x
40.06505 - 38.227 = -95.21x + 116.88x
1.83805 = 21.67x
x = 1.83805/21.67 = 0.0848 g
moles of MgCl[tex]_{2}[/tex] = 0.0848g/95.21 g/mol = 8.909*10^-4 moles
