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:
[tex]M_{1} * V_{1} = M_{2} * V_{2}[/tex]
[tex]2.50 * 0.05 = 18.0 * V_{2}[/tex]
[tex]0.125 = 18.0\:V_2[/tex]
[tex]18.0\:V_2 = 0.125[/tex]
[tex]V_2 = \dfrac{0.125}{18.0}[/tex]
[tex]V_2 \approx 0.00694\:L \to \boxed{\boxed{V_2 \approx 6.94\:mL}}\:\:\:\:\:\:\bf\green{\checkmark}[/tex]
Answer:
The volume is approximately 6.94 mL
_______________________________
[tex]\bf\green{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}[/tex]
Answer:
6.94 ml
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 = [tex]\frac{2.50 M x 50 ml}{18 M}[/tex] = 6.94 ml
