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KatRina [158]
3 years ago
13

A solution is prepared by dissolving 60.0 g of sucrose, C12H22O11, in 250. g of water at 25°C. What is the vapor pressure of the

solution if the vapor pressure of water at 25°C is 23.76 mm Hg?
Chemistry
1 answer:
Rina8888 [55]3 years ago
7 0

Answer:

23.46 mmHg is the vapor pressure for the solution

Explanation:

To solve this problem we need to apply a colligative property, which is the lowering vapor pressure.

The formula for this is: P°- P' = P° . Xm

where P' is vapor pressure for solution and P°, vapor pressure for pure solvent.

Let's determine the Xm (mole fraction for solute)

We calculate the moles of the solute and the solvent and we sum each other:

Moles of solute: 60 g /342 g/mol = 0.175 moles of sucrose

Moles of solvent: 250 g / 18 g/mol = 13.8 moles of water

Total moles: 13.8 moles + 0.175 moles = 13.975 moles

Xm for solute: 0.175 moles / 13.975 moles = 0.0125

Let's replace data in the formula: 23.76 mmHg - P' = 23.76 mmHg . 0.0125

P' = - (23.76 mmHg . 0.0125 - 23.76 mmHg) → 23.46 mmHg

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For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
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<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

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