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frez [133]
3 years ago
5

What is the molarity of a solution containing 8.0 moles of solute in 120 mL of solution

Chemistry
2 answers:
mamaluj [8]3 years ago
8 0
Molarity=\frac{moles}{Liters}

120ml= .12L

M=\frac{8.0}{.12}

M= 66.667
jeka943 years ago
5 0

Answer : The molarity of the solution is 66.7 M

Explanation :

Molarity of a solution is defined as the moles of solute dissolved in 1 liter of solution. It is denoted as M

The equation for finding molarity is given below.

M = \frac{mol}{L}

We have been given,

mol = 8.0

Volume of the solution is given in mL. Let us convert it to L.

120 mL \times \frac{1L}{1000mL}= 0.12 L

Let us plug in the above values in molarity formula.

M = \frac{8.0mol}{0.12L}

M = 66.7 mol/L

The molarity of the solution is 66.7 M

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Answer:

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All of the following reactions can be described as displacement reactions except:____________.
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Answer:

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Acid &amp; Base Calculations Calculate the hydronium ion concentration for each. Tell whether it is an acid or a base. 1. pH = 5
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<u>Explanation:</u>

pH is the negative logarithm of hydronium ion concentration present in a solution.

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  • If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14
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To calculate the pH of the solution, we use equation:

pH=-\log[H_3O^+]     ......(1)

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14                ........(2)

  • <u>For 1:</u>

We are given:

pH = 5.54

Putting values in equation 1, we get:

5.54=-\log[H_3O^+]

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Now, putting values in equation 2, we get:

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pOH = 8.46

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  • <u>For 2:</u>

We are given:

pOH = 9.7

Putting values in equation 2, we get:

14 = 9.7 + pH

pH = 4.3

Now, putting values in equation 1, we get:

4.3=-\log[H_3O^+]

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The solution is acidic in nature.

  • <u>For 3:</u>

We are given:

pH = 7.0

Putting values in equation 1, we get:

7.0=-\log[H_3O^+]

[H_3O^+]=1.00\times 10^{-7}M

Now, putting values in equation 2, we get:

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  • <u>For 4:</u>

We are given:

pH = 12.9

Putting values in equation 1, we get:

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Now, putting values in equation 2, we get:

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We are given:

pOH = 1.2

Putting values in equation 2, we get:

14 = 1.2 + pH

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Now, putting values in equation 1, we get:

12.8=-\log[H_3O^+]

[H_3O^+]=1.58\times 10^{-13}M

The solution is basic in nature.

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We are given:

[H_3O^+]=1\times 10^{-5}M

Putting values in equation 1, we get:

pH=-\log(1\times 10^{-5})

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