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KatRina [158]
2 years ago
9

If 3.53 g of CuNO, is dissolved in water to make a 0.330 M solution, what is the volume of the solution in milliliters?

Chemistry
1 answer:
solmaris [256]2 years ago
8 0

Answer:

84.8 mL

Explanation:

From the question given above, the following data were obtained:

Mass of CuNO₃ = 3.53 g

Molarity of CuNO₃ = 0.330 M

Volume of solution =?

Next, we shall determine the number of mole in 3.53 g of CuNO₃. This can be obtained as follow:

Mass of CuNO₃ = 3.53 g

Molar mass of CuNO₃ = 63.5 + 14 + (16×3)

= 63.5 + 14 + 48

= 125.5 g/mol

Mole of CuNO₃ =?

Mole = mass / Molar mass

Mole of CuNO₃ = 3.53 / 125.5

Mole of CuNO₃ = 0.028 moles

Next, we shall determine the volume of the solution. This can be obtained as follow:

Molarity of CuNO₃ = 0.330 M

Mole of CuNO₃ = 0.028 moles

Volume of solution =?

Molarity = mole /Volume

0.330 = 0.028 / Volume

Cross multiply

0.330 × Volume = 0.028

Divide both side by 0.330

Volume = 0.028 / 0.330

Volume = 0.0848 L

Finally, we shall convert 0.0848 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.0848 L = 0.0848 L × 1000 mL / 1 L

0.0848 L = 84.8 mL

Therefore, the volume of the solution is 84.8 mL.

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BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄  

Explanation:

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E/mol·L⁻¹:                   5.0 × 10⁻⁵     c

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E/mol·L⁻¹:                 2.0 × 10⁻⁷      c

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(c) Silver oxalate

                 Ag₂C₂O₄ ⇌   2Ag⁺   +   C₂O₄²⁻  

E/mol·L⁻¹:                      3.0 × 10⁻⁵       c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

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BaC₂O₄ will precipitate when   c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when   c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028       mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

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