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marishachu [46]
3 years ago
15

A student determines the iron(III) content of a solution by first precipitating it as iron(III) hydroxide, and then decomposing

the hydroxide to iron(III) oxide by heating. How many grams of iron(III) oxide should the student obtain if her solution contains 52.0 mL of 0.404 M iron(III) nitrate?
Chemistry
1 answer:
Semmy [17]3 years ago
6 0

Answer:

3.3535 g

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For iron(III) nitrate :

Molarity = 0.404 M

Volume = 52.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 52.0×10⁻³ L

Thus, moles of iron(III) nitrate :

Moles=0.404 \times {52.0\times 10^{-3}}\ moles

Moles of iron(III) nitrate  = 0.021 moles

1 mole of iron(III) nitrate forms 1 mole of iron(III) hydroxide which further forms 1 mole of  iron(III) oxide.

Thus, moles of  iron(III) oxide formed from 0.021 moles of iron(III) nitrate = 0.021 moles

Also, molar mass of iron(III) oxide = 159.69 g/mol

So, mass of iron(III) oxide = 0.021 moles × 159.69 g/mol = 3.3535 g

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<u>Answer:</u> The mass of sucrose required is 69.08 g

<u>Explanation:</u>

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Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

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T = Temperature of the solution = 290 K

Putting values in above equation, we get:

8.80atm=1\times \frac{\text{Mass of sucrose}\times 1000}{342.3\times 546}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 290K\\\\\text{Mass of sucrose}=\frac{8.80\times 342.3\times 546}{1\times 1000\times 0.0821\times 290}=69.08g

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According to
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Answer:

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