10.0 dm3 of water was added to 2.0 mol dm-3 of 2.5dm3 solution of HCl. What is the concentration of the final solution in mol dm-3?
0.4
8.0
2.0
0.5
Explanation
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Yes, my school is right
You will add the volume of water to the initial volume in order to get the final volume I.e
C1=2.0
V1=2.5
C2=?
V2=2.5+10=12.5
C1V1=C2V2
C2=C1V1/V2
C2=2.0×2.5/12.5
C2=4.5/12.5
C2=0.36 App 0.4
This is it
My school, Try breaking down a solving into simpler form, so that everybody will understand. Thanks
To solve this problem, we can use the formula:
M1V1 = M2V2
where M1 and V1 are the initial concentration and volume of the solution, and M2 and V2 are the final concentration and volume of the solution after dilution.
Initially, we have:
M1 = 2.0 mol dm^-3
V1 = 2.5 dm^3
After adding 10.0 dm^3 of water, the total volume of the solution becomes:
V2 = V1 + 10.0 dm^3 = 2.5 dm^3 + 10.0 dm^3 = 12.5 dm^3
The amount of HCl in the solution remains the same after dilution, so we can write:
M1V1 = M2V2
2.0 mol dm^-3 x 2.5 dm^3 = M2 x 12.5 dm^3
M2 = (2.0 mol dm^-3 x 2.5 dm^3) / 12.5 dm^3 = 0.4 mol dm^-3
Therefore, the concentration of the final solution is 0.4 mol dm^-3, which is option A.
i think its bcause d vol was increased by 10, as in dere was a 2.5 dm3 alredi so plus 10dm3 making 12.5...hope its clear?
The question was to find the concentration of the final solution in mol-dm3 and the formula is C1V1= C2V2 that is we are looking for C2 so C1V1÷C2 that would be 2.0 ×10÷2.5=8.0moldm-3
My school..if the ans is 0.4,then i thing the 10dm of water ned to be corrected to 12.5 dm in order to have 0.4 as the answer.bt if the 10 dm of water is use in the calculation then the ans is 0.5.pls my school do something about this question.cause are some the things that lead to some student down fall in jamb...
