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9966 [12]
2 years ago
14

What is the molar concentration of potassium ions in a 0. 250 m k2so4(aq) solution?

Chemistry
1 answer:
boyakko [2]2 years ago
7 0

The molar concentration of potassium ions in a  0. 250 m K₂SO₄(aq) solution is 2.

<h3>What is molar concentration?</h3>

The molar concentration of a solution is the presence of solute in one unit of a solution.

Molar concentration = number of moles / volume

The volume is given 0.250 m

To calculate the number of moles:

There will be 2 moles of potassium and  1 mole of sulfate in one L of the solution.

So in 0.250 L

0.250 x 2 x 1 = 0.500

The total number of moles of potassium and sulfate ion are 0.500

Now, putting the value:

Molar concentration = number of moles / volume

0.500 / 0.250 = 2

Thus, the molar concentration is 2.

To learn more about molar concentration, refer to the below link:

brainly.com/question/21841645

#SPJ4

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The solubility of CO2 in water is 0.161 g/100 mL at 20oC and a partial pressure of CO2 of 760 mmHg. What partial pressure of CO2
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<u>Answer:</u> The partial pressure of carbon dioxide having solubility 0.886g/100mL is 4182.4 mmHg

<u>Explanation:</u>

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

The equation given by Henry's law is:

C_{CO_2}=K_H\times p_{CO_2}       ......(1)

where,

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p_{CO_2} = partial pressure of carbon dioxide = 760 mmHg

Putting values in equation 1, we get:

760mmHg=K_H\times 0.161g/100mL\\\\K_H=\frac{760mmHg}{0.161g/100mL}=4720.5g.mmHg/100mL

Now, calculating the pressure of carbon dioxide using equation 1, we get:

C_{CO_2 = solubility of carbon dioxide in water = 0.886 g/100 mL

K_H = Henry's constant = 4720.5 g.mmHg/100 mL

p_{CO_2} = partial pressure of carbon dioxide = ?

Putting values in equation 1, we get:

p_{CO_2}=4720.5g.mmHg/100mL\times 0.886g/100mL=4182.4mmHg

Hence, the partial pressure of carbon dioxide having solubility 0.886g/100mL is 4182.4 mmHg

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