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Andrew [12]
3 years ago
13

Solutions A and B contain the same compound dissolved in water. Solution A is 5.0L at 0.020M and solution B is 50.0 mL at 0.60M.

How many liters of solution A are needed to contain the same number of moles as the 50.0 mL of solution B?
Chemistry
1 answer:
Otrada [13]3 years ago
7 0

<u>Answer:</u> The volume of solution A required is 0.15 L

<u>Explanation:</u>

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

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}        .......(1)

  • <u>For Solution B:</u>

Molarity of solution B = 0.60 M

Volume of solution B = 50.0 mL  = 0.050 L    (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

0.60M=\frac{\text{Moles of solution B}}{0.050L}\\\\\text{Moles of solution B}=(0.60mol/L\times 0.050L)=0.003mol

Now, calculating the volume of solution A required for the same number of moles as solution B.

Moles of solution A = 0.003 moles

Molarity of solution A = 0.020 M

Putting values in equation 1, we get:

0.020M=\frac{0.003mol}{\text{Volume of solution A}}\\\\\text{Volume of solution A}=\frac{0.003}{0.02}=0.15L

Hence, the volume of solution A required is 0.15 L

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