Question

When solutions of silver nitrate and sodium chloride are mixed, silver chloride precipitates out of the solution according to the equation:
AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq)

mass of AgCl =46.6 g

The reaction described in Part A required 3.95L of sodium chloride. What is the concentration of this sodium chloride solution?

Answer 1

The concentration of sodium chloride would be 0.082 M.

Define the molarity of a solution.

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

Stoichiometric calculations

From the equation of the reaction,

$AgNO_3(aq)+NaCl(aq)$ → $AgCl(s)+NaNO_3(aq)$

The ratio of AgCl produced to NaCl required is 1:1.

Mole of 46.6 g AgCl produced = $\frac{46.6}{43.32}$ = 0.325 moles

Equivalent mole of NaCl = 0.325 moles.

Molarity of 0.325 moles, 3.95 L NaCl

Molarity = $\frac{Moles \;solute}{Volume}$

Molarity =$\frac{0.325}{3.95}$

Molarity = 0.082 M

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Answer 2

The concentration of the sodium chloride would be 0.082 M

Stoichiometric calculations

From the equation of the reaction, the ratio of AgCl produced to NaCl required is 1:1.

Mole of 46.6 g AgCl produced = 46.6/143.32 = 0.325 moles

Equivalent mole of NaCl = 0.325 moles.

Molarity of 0.325 moles, 3.95 L NaCl = mole/volume = 0.325/3.95 = 0.082 M

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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