A sample containing 0.2784 grams of sodium chloride (NaCl) and 0.5486 grams of m
ID: 789374 • Letter: A
Question
A sample containing 0.2784 grams of sodium chloride (NaCl) and 0.5486 grams of magnesium chloride (MgCl2). The chloride in the sample was precipitated by the addition of 47.8 mL of a silver nitrate solution. What is the concentration of the silver nitrate solution?A sample containing 0.2784 grams of sodium chloride (NaCl) and 0.5486 grams of magnesium chloride (MgCl2). The chloride in the sample was precipitated by the addition of 47.8 mL of a silver nitrate solution. What is the concentration of the silver nitrate solution?
Explanation / Answer
moles of cl in Nacl= 0.2784/58.5= 4.7589 millimoles
moles of cl in Mgcl2= 2* 0.5486/95= 11.549 millimoles
total cl= 16.3079 millimoles= moles of Ag
concentration= 16.3079/47.8= 0.3411 M