Question

In a schools laboratory students require 50.0 mL of 2.50 H2SO4 for an experiment but only available stock solution of the acid has a concentration of 18.0m. What volume of the stock would they use to make the required solution

0.900Ml
1.11ml
6.94 ml
7.20ml

Answer 1

Hello!

In a schools laboratory students require 50.0 mL of 2.50 H2SO4 for an experiment but only available stock solution of the acid has a concentration of 18.0m. What volume of the stock would they use to make the required solution

0.900 mL

1.11 mL

6.94 mL

7.20 mL

We have the following data:

M1 (initial molarity) = 2.50 M (or mol/L)

V1 (initial volume) = 50.0 mL → 0.05 L

M2 (final molarity) = 18.0 M (or mol/L)

V2 (final volume) = ? (in mL)

Let's use the formula of dilution and molarity, so we have:

`M_(1) * V_(1) = M_(2) * V_(2)`

`2.50 * 0.05 = 18.0 * V_(2)`

`0.125 = 18.0\:V_2`

`18.0\:V_2 = 0.125`

`V_2 = (0.125)/(18.0)`

`V_2 \approx 0.00694\:L \to \boxed{\boxed{V_2 \approx 6.94\:mL}}\:\:\:\:\:\:\bf\green{\checkmark}`

Answer:

The volume is approximately 6.94 mL

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Answer 2

6.94 ml

Step-by-step explanation:

Using the dilution equation,

C₁V₁ = C₂V ₂

where C is the concentration in molarity and V is the volume in ml

C₁= 2.50 M, V₁ = 50 ml, C₂ = 18 M

2.50 M x 50 ml = 18 M x V₂

V₂ =
`(2.50 M x 50 ml)/(18 M)` = 6.94 ml