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jeyben [28]
3 years ago
7

In an experiment, a 88.11 mL sample of unknown silver nitrate solution was treated with 9.753 g of sodium chloride, resulting in

4.576 g of precipitate. Calculate the molarity of the silver nitrate solution
Chemistry
1 answer:
zlopas [31]3 years ago
4 0

Answer:

0.362 M

Explanation:

The reaction that takes place is:

  • AgNO₃ + NaCl → NaNO₃ + AgCl(s)

First we <u>convert the mass of AgCl (the precipitate) into moles</u>, using its <em>molar mass</em>:

  • 4.576 g AgCl ÷ 143.32 g/mol = 0.0319 mol AgCl

Now we <u>convert AgCl moles into AgNO₃ moles:</u>

  • 0.0319 mol AgCl * \frac{1molAgNO_3}{1molAgCl} = 0.0319 mol AgNO₃

Finally we <u>calculate the molarity</u>, after <em>converting 88.11 mL to L</em>:

  • 88.11 mL / 1000 = 0.08811 L
  • Molarity = 0.0319 mol AgNO₃ / 0.08811 L = 0.362 M
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<h3><u>Answer;</u></h3>

Atomic particles

The first step in the two-step process of making a solution is the breakdown of the solute source into <u>atomic particles</u>.

<h3><u>Explanation</u>;</h3>
  • Solution is a homogeneous mixture of two or more substances uniformly dispersed throughout a single phase
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1.62 × 10²⁴ atoms are in 52.3 g of lithium hypochlorite.

Explanation:

To find the amount of atoms that are in 52.3 g of lithium hypochlorite, we must first find the amount of moles. We do this by dividing by the molar mass of lithium hypochlorite.

52.3 g ÷ 58.4 g/mol = 0.896 mol

Next we must find the amount of formula units, we do this be multiplying by Avagadro's number.

0.896 mol × 6.02 × 10²³ = 5.39 × 10²³ f.u.

Now to get the amount of atoms we can multiply the amount of formula units by the amout of atoms in one formula unit.

5.39 × 10²³ f.u. × 3 atom/f.u. = 1.62 × 10²⁴ atoms

1.62 × 10²⁴ atoms are in 52.3 g of lithium hypochlorite.

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