What volume of (dm\(^3\)) of water will be added to 10dm\(^3\) of 2.0 mol/dm\(^3\) HCL acid solution to give a final solution of 0.5 mol/dm\(^3\)?
30
40
20
50
Explanation
Video Explanation
Post your Contribution
Discussions (15)
the question says what volume of water will be added,remember that the solution had an initial vol. of 10dm3,therefore
vol. of water added=final vol of the solution -initial volume of the solution
c1v1=c2v2
10×2=0.5×v2
=40dm3
vol. of water added to make the new volume=40-10=30dm3
The correct answer is 30cm3 not 40cm3 because when you find the final volume you have to subtract it from the initial V1 because they asked for the volume that needs to be added to the initial.
I think that the final volume of 40cm3 which was gotten as the answer, the 10cm3 of HCl needs to be substracted from the 40cm3 to give 30cm3 which is the volume of the water added not the final volume of the solution
Myschool, The correct answer is 30, but the video explanation is not complete as it stops at 40.
Please correct it, or remove the video 👍👍
My school didn't finish up the solving...
Since u've gotten the vol of water to be 40dm3
To get the vol of the 40dm3 of water that will be added to the acid to give the final solution is V2-V1
Which is 40-10 = 30dm3
We can solve this using the dilution formula:
C_1 V_1 = C_2 V_2
Where:
(initial concentration)
(initial volume)
(final concentration)
(final volume)
Plug in the values:
2.0 \times 10 = 0.5 \times V_2
\Rightarrow 20 = 0.5 \times V_2
\Rightarrow V_2 = \frac{20}{0.5} = 40\ \text{dm}^3
To find the volume of water added:
V_{\text{added}} = V_2 - V_1 = 40 - 10 = 30\ \text{dm}^3
Answer: A. 30
