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Trail_Builder
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I thought I knew moles inside out, but I havnt quite been told how to factor in what happens when you dissolve something in something.
1.500g of a sample of limestone was dissolved in 50cm^3 of a 1.00 mol/dm^3 of hydrochloric acid solution. The resulting solution was made up to 250cm^3 precisely with distilled water. 25.0cm^3 of this solution required 21.05cm^3 of 0.100 mol/dm^3 of sodium hydroxide for neutralisation.
Calculate the % of rock which is calcium carbonate.
mol of HCL:
mol = (vol/1000)*conc
50cm^3 is 1 mol/dm^3, so added with water up to 250cm^3 will be 0.2 mol/dm^3
mol(HCl) = 0.250*0.2
mol(HCl) = 0.05
but then we only take 25cm^3, so...
mol(HCl) = 0.005
mol of NaOH:
mol(NaOH) = (vol/1000)*conc
mol(NaOH) = (21.05/1000)*0.1
mol(NaOH) = 0.002105
Now I get stuff, I am not sure what the effect of dissolving 1.5g of limestone will even have, so I am stuck :S.
Hope you can help.
Thanks
Homework Statement
1.500g of a sample of limestone was dissolved in 50cm^3 of a 1.00 mol/dm^3 of hydrochloric acid solution. The resulting solution was made up to 250cm^3 precisely with distilled water. 25.0cm^3 of this solution required 21.05cm^3 of 0.100 mol/dm^3 of sodium hydroxide for neutralisation.
Calculate the % of rock which is calcium carbonate.
Homework Equations
The Attempt at a Solution
mol of HCL:
mol = (vol/1000)*conc
50cm^3 is 1 mol/dm^3, so added with water up to 250cm^3 will be 0.2 mol/dm^3
mol(HCl) = 0.250*0.2
mol(HCl) = 0.05
but then we only take 25cm^3, so...
mol(HCl) = 0.005
mol of NaOH:
mol(NaOH) = (vol/1000)*conc
mol(NaOH) = (21.05/1000)*0.1
mol(NaOH) = 0.002105
Now I get stuff, I am not sure what the effect of dissolving 1.5g of limestone will even have, so I am stuck :S.
Hope you can help.
Thanks
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