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solong [7]
3 years ago
6

How would you prepare a 0.1M solution of HCl starting with a 1.0M solution? Assume you want to prepare 100 ml of solution.

Chemistry
1 answer:
omeli [17]3 years ago
4 0
This question is a dilution problem. In order to prepare a certain concentration and volume from a stock solution, you have to use the equation,

M1V1 = M2V2

From the equation, you need to calculate the volume you need from the stock solution to be able to have the diluted concentration then dilute it to a certain volume.

(1.0 M) (V1) = (0.1 M) (100 mL)
V1 = 0.1 mL is needed from the 1.0 M HCl solution

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A solution of 2-propanol and 1-octanol behaves ideally. Calculate the chemical potential of 2-propanol in solution relative to t
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Answer:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

Explanation:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:

\mu (l) = \mu ^{\circ} (l) + R*T*ln(x)

<u>Where:</u>

<em>μ (l): is the chemical potential of 2-propanol in solution    </em>

<em>μ° (l): is the chemical potential of pure 2-propanol   </em>

<em>R: is the gas constant = 8.314 J K⁻¹ mol⁻¹ </em>

<em>T: is the temperature = 82.3 °C = 355.3 K </em>

<em>x: is the mole fraction of 2-propanol = 0.41 </em>

\mu (l) = \mu ^{\circ} (l) + 8.314 \frac{J}{K*mol}*355.3 K*ln(0.41)

\mu (l) = \mu ^{\circ} (l) - 2.63 \cdot 10^{3} J*mol^{-1}

\mu (l) - \mu ^{\circ} (l) = - 2.63 \cdot 10^{3} J*mol^{-1}  

Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

I hope it helps you!    

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