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Paladinen [302]
3 years ago
13

What is the molarity of a 3.0-liter solution that contains 0.45 moles of solute

Chemistry
2 answers:
sergeinik [125]3 years ago
6 0
Moles = n/v where n is the moles of solute and v being the liters of solution.
We can put in the information provided to find the molarity.

Moles = .45/3.0 = .15
So we now know that the molarity of that solution is .15!
 I hope I helped you :). Make sure to memorize that formula because it's not that hard as long as you know what to plug in.
Ksivusya [100]3 years ago
5 0

Answer : The molarity of the solution is, 0.15 mole/L

Solution :

Molarity : It is defined as the number of moles of solute present in one liter of the solution.

Formula used :

Molarity=\frac{\text{moles of solute}}{\text{Volume of solution in liters}}

Now put all the given values in this formula, we get the molarity of the solution.

Molarity=\frac{0.45mole}{3L}=0.15mole/L

Therefore, the molarity of the solution is, 0.15 mole/L

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Hope this helps!
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Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t
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Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

A=\epsilon \times C\times l

A=\log \frac{I_o}{I}

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where,

A = absorbance of solution

C = concentration of solution = 3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}

l = path length = 2.5 mm = 0.25 cm

I_o = incident light

I = transmitted light

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Now put all the given values in the above formula, we get:

\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)

\log \frac{I_o}{I}=0.6947

\frac{I_o}{I}=10^{0.6947}=4.951

If we consider I_o = 100

then, I=\frac{100}{4.951}=20.198

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light I_A = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100

\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%

Therefore, the percentage reduction in intensity is 79.80 %

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