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?

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

Answer 1

[AI-generated answer — Bot GPT 3.5]

To determine the concentration of the silver nitrate solution, we need to calculate the moles of chloride ions in the sample and then use that information to find the concentration of the silver nitrate solution.

1. Calculate the moles of NaCl and MgCl2:
- Moles of NaCl = 0.2784 g / 58.44 g/mol = 0.00476 mol
- Moles of MgCl2 = 0.5486 g / (24.31 + 2*35.45) g/mol = 0.00473 mol

2. Calculate the total moles of chloride ions:
Total moles of chloride ions = Moles of NaCl + 2 x Moles of MgCl2
Total moles of chloride ions = 0.00476 + 2 x 0.00473 = 0.01422 mol

3. Calculate the volume of the silver nitrate solution in liters:
Volume of silver nitrate solution = 47.8 mL = 0.0478 L

4. Use the balanced chemical equation to determine the mole ratio between silver nitrate and chloride ions:
AgNO3 + Cl- -> AgCl + NO3-
From the balanced equation, it is clear that 1 mole of AgNO3 reacts with 1 mole of Cl-

5. Calculate the concentration of the silver nitrate solution:
Concentration of AgNO3 = Moles of Cl- / Volume of silver nitrate solution
Concentration of AgNO3 = 0.01422 mol / 0.0478 L = 0.298M AgNO3

Therefore, the concentration of the silver nitrate solution is approximately 0.298M AgNO3. So, the closest option in the list provided is (a) 0.341M AgNO3.