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Lerok [7]
2 years ago
12

|

Chemistry
1 answer:
lilavasa [31]2 years ago
6 0

1.66 M is the concentration of the chemist's working solution.

<h3>What is molarity?</h3>

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.

In this case, we have a solution of Zn(NO₃)₂.

The chemist wants to prepare a dilute solution of this reactant.

The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.

We want to know the concentration of this diluted solution.

As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:

n_1= n_2  (1)

and we also know that:

n = M x V_2

If we replace this expression in (1) we have:

M_1 x V_1= M_2 x V_2

Where 1, would be the stock solution and 2, the solution we want to prepare.

So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is M_2:

4.93 x 210 =  620 xM_2

M_2 = 1.66 M

This is the concentration of the solution prepared.

Learn more about molarity here:

brainly.com/question/19517011

#SPJ1

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What mass of precipitate forms when 185.5 ml of 0.533 m naoh is added to 627 ml of a solution that contains 15.8 g of aluminum s
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2,57 g of precipitate.

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For the reaction:

6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄

The precipitate is Al(OH)₃.

185,5mL of 0,533M NaOH are:

0,1855L × 0,533M = <em>0,0989 moles NaOH</em>

Moles of Al₂(SO₄)₃ are:

15,8g × \frac{1mol}{342,15g} = <em>0,0462 moles Al₂(SO₄)₃</em>

For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:

0,0989moles NaOH × \frac{1molAl_{2}(SO_{4})_{3}}{6 moles NaOH} = <em>0,0165 moles Al₂(SO₄)₃</em>

As you have <em>0,0462 moles Al₂(SO₄)₃ </em>the limiting reactant is NaOH.

0,0989 moles of NaOH produce:

0,0989moles NaOH × \frac{2molAl(OH)_{3}}{6 moles NaOH} = <em>0,0330 moles of Al(OH)₃</em>

These moles are:

0,0330 moles of Al(OH)₃ × (78 g/mol) = <em>2,57 g of Al(OH)₃ ≡ mass of precipitate</em>

<em></em>

I hope it helps!

<em> </em>

3 0
3 years ago
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