English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

We have a 10% aqueous solution and have to replace it. However, I can only find the solution in 0.01N, 0.025N, 0.0375N, 0.1N and 1.0N. Which one is the right one? and Why?

2006-12-14 23:40:14 · 4 answers · asked by Lopinion 1 in Science & Mathematics Chemistry

4 answers

A solution of what? If you don't specify the compound we can't say.

Usually 10% means 10 g of solute per 100 ml of solution.
If the solute is liquid then usually it means 10 ml of solute per 100 ml of solution.Then you would need the density of the solute to convert into g/100ml.
(Sometimes it is w/w meaning 10 g of solute per 100 g of solution but I doubt if that is the case here.If it is you need the density of the solution so that you can convert the mass of the solution into volume)

Multiply by 10 to have g solute/L solution and then divide by the equivalent weight to get the normality.
The equivalent weight is the molecular weight divided by the number of chemical units per molecule for the reaction you are interested (e.g. for acid and bases it is the number of H+ or OH- per molecule respectively, for oxidants/reductants the number of electrons gained/lost, etc)

2006-12-15 02:50:38 · answer #1 · answered by bellerophon 6 · 0 1

It would depend if it is 10% by mass or by moles. If it is by moles then the correct answer is 0.1N. 10% in decimal form is 0.1.

2006-12-14 23:50:55 · answer #2 · answered by j_son_06 5 · 0 0

1.oN

2006-12-14 23:43:44 · answer #3 · answered by George 2 · 0 2

one million.) .0204Lx(.45moles/L)x(2moles OH-/1mole Ba(OH)2)x(1mole H+/1mole OH-)x(1mole HClO4/1moleH+)x(one hundred.5g/mole)= one million.845g (one million.845g of solute/12.1g of answer)= 15.2% HClO4 2.) .0387mLx(.514moles/L)=.01989 moles of HNO3 .027Lx(.411moles/L)=.01110 moles of KOH We assume it reacts completly like in a titration. .01989moles-.01110moles=.008795moles .008795moles of HNO3 left x(1mole H+/1mole HNO3)x(1mole OH-/1moleH+)x(1mole Ca(OH)2/2moles OH-)= .004397moles of Ca(OH)2 (.004397moles/.0207L)= .2124M Ca(OH)2 3.)one million.01gx(mole/204g)x(1mole KOH/1mole PH)= .004951moles of KOH (.004951moles/.0365L)=.1356M KOH .0221L of KOH x (.1356 moles/L)x(1mole HCl/1mole KOH)= .002997moles of HCl (.002997moles/.0219L)= .1369M HCl

2016-12-30 11:14:39 · answer #4 · answered by Anonymous · 0 0

fedest.com, questions and answers