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levacccp [35]
3 years ago
11

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:
svet-max [94.6K]3 years ago
3 0

Answer:

M=0.362M

Explanation:

Hello!

In this case, according to the following chemical reaction:

AgNO_3(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)

It is possible to compute the moles of silver nitrate via stoichiometry that produced 4.576 g of silver chloride as shown below:

n_{AgNO_3}=4.576gAgCl*\frac{1molAgCl}{143.32gAgCl}*\frac{1molAgNO_3}{1molAgCl}\\\\n_{AgNO_3}=0.03193molAgNO_3

Thus, since the molarity is obtained by dividing moles by volume, we obtain:

M=\frac{0.03193mol}{0.08811L}\\\\M=0.362M

Best regards!

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A 2.300×10−2 m solution of nacl in water is at 20.0∘c. the sample was created by dissolving a sample of nacl in water and then b
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The given question is incomplete. The complete question is as follows.

A 2.300×10−2 m solution of nacl in water is at 20.0∘c. the sample was created by dissolving a sample of nacl in water and then bringing the volume up to 1.000 l. it was determined that the volume of water needed to do this was 999.4 ml . the density of water at 20.0∘c is 0.9982 g/ml.

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Molality is defined as the number of moles present in kg of a solvent.

Mathematically,     Molality = \frac{\text{moles of solute}}{\text{mass of solvent}}

Also,

      Mole of solute = Molarity of solute x Volume of solution

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Therefore, mass of solvent will be as follows.

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Therefore, we will calculate the molality as follows.

          Molality = \frac{0.0230 mol}{0.9977 kg}

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