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You have a stock solution of 15.6 M NH3. How many milliliters of this solution should you dilute to make 1450 mL of 0.210 M NH3?If you take a 15.0-mL portion of the stock solution and dilute it to a total volume of 0.600 L , what will be the concentration of the final solution?

Respuesta :

Answer:

For 1: 19.52mL of the stock solution must be taken.

For 2: The concentration of the final solution will be 0.39M

Explanation:

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

[tex]M_1V_1=M_2V_2[/tex]

where, [tex]M_1\text[ and }V_1[/tex] are the molarity and volume of one solution

[tex]M_2\text[ and }V_2[/tex] are the molarity and volume of another solution

  • For 1:

[tex]M_1[/tex] = Molarity of the stock solution = 15.6M

[tex]V_1[/tex] = Volume of the stock solution = ? mL

[tex]M_2[/tex] = Molarity of the diluted solution = 0.210M

[tex]V_2[/tex] = Volume of the diluted solution = 1450 mL

Putting values in above equation, we get:

[tex]15.6\times V_1=0.210\times 1450\\\\V_1=19.52mL[/tex]

  • For 2:

[tex]M_1[/tex] = Molarity of the stock solution = 15.6M

[tex]V_1[/tex] = Volume of the stock solution = 15 mL

[tex]M_2[/tex] = Molarity of the diluted solution = ? M

[tex]V_2[/tex] = Volume of the diluted solution = 0.600L = 600mL    (Conversion factor: 1L = 1000mL)

Putting values in above equation, we get:

[tex]15.6\times 15=\times 1450\\\\M_2=0.39M[/tex]

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