Question

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?

Answer 1

To make 1450 mL of 0.210 M NH3, dilute 19.5 mL of a 15.6 M NH3 stock solution. For a final volume of 0.600 L from a 15.0-mL portion of the same stock, the final concentration will be 0.390 M NH3.

Step-by-step explanation:

To determine how many milliliters of a 15.6 M NH3 stock solution you need to dilute to make 1450 mL of a 0.210 M NH3 solution, you can use the dilution equation (C1V1 = C2V2), where C1 is the concentration of the stock solution, V1 is the volume of stock solution needed, C2 is the desired concentration, and V2 is the final volume of the diluted solution.

Using the equation:

15.6 M (V1) = 0.210 M (1450 mL)

V1 = (0.210 M * 1450 mL) / 15.6 M

V1 = 19.5 mL

Therefore, you need to dilute 19.5 mL of the stock solution to make 1450 mL of a 0.210 M NH3 solution.

For the second part, if you take a 15.0-mL portion of the stock solution and dilute it to a total volume of 0.600 L, the final concentration (C2) can be found using the same dilution equation:

15.6 M (15.0 mL) = C2 (600 mL)

C2 = (15.6 M * 15.0 mL) / 600 mL

C2 = 0.390 M

The concentration of the final solution will be 0.390 M NH3.

Answer 2

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

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

Step-by-step explanation:

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

`M_1V_1=M_2V_2`

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

`M_2\text[ and }V_2` are the molarity and volume of another solution

• For 1:

`M_1` = Molarity of the stock solution = 15.6M

`V_1` = Volume of the stock solution = ? mL

`M_2` = Molarity of the diluted solution = 0.210M

`V_2` = Volume of the diluted solution = 1450 mL

Putting values in above equation, we get:

`15.6* V_1=0.210* 1450\\\\V_1=19.52mL`

• For 2:

`M_1` = Molarity of the stock solution = 15.6M

`V_1` = Volume of the stock solution = 15 mL

`M_2` = Molarity of the diluted solution = ? M

`V_2` = Volume of the diluted solution = 0.600L = 600mL (Conversion factor: 1L = 1000mL)

Putting values in above equation, we get:

`15.6* 15=* 1450\\\\M_2=0.39M`