Question

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?
*
1 point
(a) 0.341M AgNO3
(b) 0.220 M AgNO3
(c) 0.00341 M AgNO3
non of all

Answer 1

To find the concentration of the silver nitrate solution, we first need to determine the moles of chloride ions in the sample, as they will react with the silver nitrate to form a precipitate.

1. Moles of NaCl:
Molar mass of NaCl = 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol
Moles of NaCl = 0.2784 g / 58.44 g/mol = 0.00476 mol

2. Moles of MgCl2:
Molar mass of MgCl2 = 24.31 g/mol (Mg) + 2 * 35.45 g/mol (Cl) = 95.21 g/mol
Moles of MgCl2 = 0.5486 g / 95.21 g/mol = 0.00577 mol

Total moles of chloride ions = 0.00476 mol (from NaCl) + 0.01154 mol (from MgCl2) = 0.01053 mol

Since the silver nitrate reacts with chloride ions in a 1:1 ratio, the moles of AgNO3 needed to react with all the chloride ions in the sample is equal to 0.01053 mol.

3. Volume of silver nitrate solution = 47.8 mL = 0.0478 L

Concentration of AgNO3 = Moles of AgNO3 / Volume of solution
Concentration of AgNO3 = 0.01053 mol / 0.0478 L = 0.220 M AgNO3

Therefore, the concentration of the silver nitrate solution is 0.220 M AgNO3, which corresponds to option (b).