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Olenka [21]
3 years ago
12

how does the percentage by mass of the solute describe the concentration of an aqueous solution os potassium sulfate

Chemistry
1 answer:
kirza4 [7]3 years ago
3 0

Answer:

Explanation:

In an aqueous solution of potassium sulfate (K₂SO₄), the solute is K₂SO₄ and the solvent is water. The percentage by mass describes the grams of solute there are dissolved per 100 grams of solution. It can be calculated as:

mass percentage = (mass of solute/total mass of solution) x 100%

For example, in an aqueous solution which is 2% by mass of K₂SO₄, there are 2 grams of K₂SO₄ per 100 g of solution.

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Please help me.<br> Thank you!
Katarina [22]

Answer: Positive

Explanation: Just a guess

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3 years ago
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Aspirin is a weak organic acid whose molecular formula is HC9H7O4. An aqueous solution of aspirin is prepared by dissolving 3.60
garri49 [273]

Answer:

Ka = 3.50x10⁻⁴

Explanation:

First, we need to convert the unit of 3.60 g/L to mol/L:

C_{C_{9}H_{8}O_{4}} = 3.60 \frac{g}{L}*\frac{1 mol}{180.16 g} = 0.0200 mol/L

The reaction dissociation of aspirin in water is:

C₉H₈O₄  +  H₂O  ⇄  C₉H₇O₄⁻ + H₃O⁺    

0.02 - x                        x             x

The constant of the above reaction is:

Ka = \frac{[C_{9}H_{7}O_{4}^{-}][H_{3}O^{+}]}{[C_{9}H_{8}O_{4}]}

Ka = \frac{x^{2}}{0.02 - x}

To find Ka we need to find the value of x. We know that pH = 2.6 so:

pH = -log[H_{3}O^{+}]

2.6 = -log(x)

x = 2.51 \cdot 10^{-3} M = [H_{3}O^{+}] = [C_{9}H_{7}O_{4}^{-}]

Now, the concentration of C₉H₈O₄ is:

C_{C_{9}H_{8}O_{4}} = 0.02 - 2.51 \cdot 10^{-3} = 0.018 M

Finally, Ka is:

Ka = \frac{[C_{9}H_{7}O_{4}^{-}][H_{3}O^{+}]}{[C_{9}H_{8}O_{4}]} = \frac{(2.51 \cdot 10^{-3})^{2}}{0.018} = 3.50 \cdot 10^{-4}

Therefore, the Ka of aspirin is 3.50x10⁻⁴.

       

I hope it helps you!

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