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標題:

chem concentration

發問:

220.0 cm3 of water are added to 10.5mol dm-3 hydrochloric acid so that the new molarity of the solution is 1.26mol dm-3. What is the volume of the original acid used?A 18.3 cm3B 25.0 cm3C 26.4 cm3D 30.0 cm3. 420.0 cm3 of water are added to 30.0 cm3 of sodium hydroxide solution so that the... 顯示更多 220.0 cm3 of water are added to 10.5mol dm–3 hydrochloric acid so that the new molarity of the solution is 1.26mol dm–3 . What is the volume of the original acid used? A 18.3 cm3 B 25.0 cm3 C 26.4 cm3 D 30.0 cm3 . 420.0 cm3 of water are added to 30.0 cm3 of sodium hydroxide solution so that the new molarity of he solution is 0.210 mol dm–3 . What is the molarity of the original solution? A 2.94mol dm–3 B 3.15mol dm–3 C 6.00mol dm–3 D 6.43mol dm–3

最佳解答:

1)The answer is D Let x cm^3 and n be the original volume and no. of moles of HCl respectively n/(x/1000) = 10.5...........................(1) n/[(x/1000)+(220/1000)] = 1.26......(2) solving (1) and (2), x=30 (n=0.315) <---- can be omitted alternative method (x/1000)(10.5)=[(x+220)/1000](1.26) x=30 2)The answer is B Let m mol dm^-3 be the molarity of original NaOH m(30/1000) = 0.21[(30+420)/1000] m=3.15 Additional information: The equation is M1V1=M2V2 Where M1, M2 ,V1, V2 are the original and new molarities, and original and new volumes. It is because the no. of moles of the solution during dilution is unchanged.

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