Question

What volume of an 18.0 M H2SO4 solution is needed to prepare 150.0mL of a 0.250 M solution

Answer 1

To prepare 150.0 mL of a 0.250 M solution, you would need approximately 0.002083 L of an 18.0 M H2SO4 solution.

Step-by-step explanation:

To calculate the volume of an 18.0 M H2SO4 solution needed to prepare 150.0 mL of a 0.250 M solution, we can use the formula:

Volume_1 * Molarity_1 = Volume_2 * Molarity_2

Plugging in the given values:

Volume_1 * 18.0 M = 150.0 mL * 0.250 M

Converting the mL to L:

Volume_1 * 18.0 M = 0.150 L * 0.250 M

Solving for Volume_1:

Volume_1 = (0.150 L * 0.250 M) / 18.0 M = 0.002083 L

Therefore, the volume of the 18.0 M H2SO4 solution needed to prepare 150.0 mL of a 0.250 M solution is approximately 0.002083 L.

Answer 2

`\huge{ \boxed{2.08 \: ml}}`

Step-by-step explanation:

The volume required can be found by using the formula:

`C_1V_1 = C_2V_2`

where

C1 is the concentration of the stock solution

V1 is the volume of the stock solution

C2 is the concentration of the final solution

V2 is the volume of the final solution

From the question we're finding V1 so we make it the subject that's:

`V_1= (C_2V_2)/(C_1) \\`

From the question:

C1 = 18 M

C2= 0.250 M

V2= 150 mL

V1=?

`\therefore \: V_1 = (0.25 * 150)/(18) \\ = 2.08 \: ml`