If 20 cm3 of distilled water is added to 80 cm3 of 0.50 mol dm-3 hydrochloric acid, the concentration of the acid will change to
20 mol dm-3
0.40 mol dm-3
2.00 mol dm-3
5.00 mol dm -3
Explanation
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When you mix 20 cm3 of distilled water with 80 cm3 of 0.50 mol dm-3 hydrochloric acid, you can calculate the final concentration of the acid using the formula:
C1V1 = C2V2
Where:
C1 = initial concentration of the acid
V1 = initial volume of the acid
C2 = final concentration of the acid
V2 = final volume of the acid (initial volume + volume of distilled water)
In this case:
C1 = 0.50 mol dm-3
V1 = 80 cm3
V2 = 80 cm3 + 20 cm3 = 100 cm3
Now, plug these values into the formula to find the final concentration (C2):
0.50 mol dm-3 * 80 cm3 = C2 * 100 cm3
40 = 100C2
C2 = 40 / 100
C2 = 0.40 mol dm-3
Therefore, when you mix 20 cm3 of distilled water with 80 cm3 of 0.50 mol dm-3 hydrochloric acid, the concentration of the acid will change to 0.40 mol dm-3.
hope this was helpful??🥲🥲
Dewizo 
🧘♂️🧘♂️
please note that you were not asked for the concentration of water...
if that were to be the case
m1v1 = m2v2
v2 = 80×0.5/20 = 2mol/dm³
rather we were asked for the connection of thediluted solution of HCL
remember the original volume of HCl was 80cm³ and it had a fixed concentration of 0.5mol/dm³..
after adding 20cm³ of water the total volume of dilute HCL now became 100cm³ and definitely the concentration had to change..(jamb is now asking "what is that new concentration"??)
C1V1 = C2V2
C2 = c1v1/v2 = 80×0.5/100 = 0.4mol/dm³
I think the question and the solution are wrongly constructed because we were asked to look for the conc.of the acid but in the question it is already given as 0.50 so I think we are supposed to look for the con.in mol/dm3 of the distilled water which is 80*0.50/20=C
V1=80cm3 M1=0.50mol/dm3
V2=100cm3 M2=unknown
Formula
M1V1=M2V2
The answer is 0.4moldm3
