Question

in a school’s laboratory, students require 50.0 mL of 2.50 M H2SO4 for an experiment, but the only available stock solution of the acid has a concentration of 18.0 M. What volume of the stock solution would they use to make the required solution? Use mc018-1.jpg.

Answer 1

Hello!

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

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:

$M_{1} * V_{1} = M_{2} * V_{2}$

$2.50 * 0.05 = 18.0 * V_{2}$

$0.125 = 18.0\:V_2$

$18.0\:V_2 = 0.125$

$V_2 = \dfrac{0.125}{18.0}$

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

Answer:

The volume is approximately 6.94 mL

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Answer 2

Ans: Volume of stock H2SO4 required = 6.94 ml

Given:

Concentration of stock H2SO4 solution M1 = 18.0 M

Concentration of the final H2SO4 solution needed M2 = 2.50 M

Final volume of H2SO4 needed, V2 = 50.0 ml

To determine:

Volume of stock needed, V1

Explanation:

Use the dilution relation:

$M1V1 = M2V2\\\\V1 = \frac{M2V2}{M1} \\\\V1 = \frac{2.50 M * 50.0 ml}{18.0 M} = 6.94 ml$