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Aneli [31]
3 years ago
6

How many grams of Cu(OH)2 will precipitate when excess NaOH solution is added to 46.0 mL of 0.584 M CuSO4

Chemistry
1 answer:
Slav-nsk [51]3 years ago
3 0
<h3>Answer:</h3>

2.624 g

<h3>Explanation:</h3>

The equation for the reaction is given as;

  • CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
  • Volume of CuSO₄ as 46.0 mL;
  • Molarity of CuSO₄ as 0.584 M

We are required to calculate the mass of Cu(OH)₂ precipitated

  • We are going to use the following steps;
<h3>Step 1: Calculate the number of moles of CuSO₄ used</h3>

Molarity = Number of moles ÷ Volume

To get the number of moles;

Moles = Molarity × volume

          = 0.584 M × 0.046 L

          = 0.0269 moles

<h3>Step 2: Calculate the number of moles of Cu(OH)₂ produced </h3>
  • From the equation 1 mole of CuSO₄ reacts to give out 1 mole of Cu(OH)₂
  • Therefore; Mole ratio of CuSO₄ to Cu(OH)₂ is 1 : 1.

Thus, Moles of CuSO₄ = Moles of Cu(OH)₂

Hence, moles of Cu(OH)₂ = 0.0269 moles

<h3>Step 3: Calculate the mass of Cu(OH)₂</h3>

To get mass we multiply the number of moles with the molar mass.

Mass = Moles × Molar mass

Molar mass of Cu(OH)₂ is 97.561 g/mol

Therefore;

Mass of Cu(OH)₂ = 0.0269 moles × 97.561 g/mol

                           = 2.624 g

Thus, the mass of Cu(OH)₂ that will precipitate is 2.624 g

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Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres
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<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

K_f = molal freezing point elevation constant = 1.86°C/m

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M_{solute} = Molar mass of solute (KCl) = 74.55 g/mol

W_{solvent} = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC

Hence, the freezing point of solution is -0.454°C

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2 years ago
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