The normality of 10 ml of a 20 V H2O2 solution is ?
@ParthKohli
do u know what's 20 V H2O2 ?
20 M H2O2? Not sure what 20 V is... Maybe volume/volume percentage?
ok..wiat !
so that means that 20 ml of H2O2 in 100 ml of solution..ryt ?
If V means %V/V, then sure.
ok let us suppose it is %v/v ....then how to proceed towards Normality ?
How to do that without any density given ?
Not sure how...
Density is required. Without that, not sure at all.
What if it is 20 molar H2O2 solution?
then it means that 20 moles in 1 litre
20 moles=34*20=680 g
Normality = Molarity * n-factor
yes..what will be the n factor here ?
@Abmon98 what do u think ?
I'm sorry, I haven't really studied this outside acids and bases. I'm Googling to see what the n-factor is in these cases.
oh i see..it's ok...thanx for the concern buddy !
btw the answer is 3.58
@abb0t @Australopithecus
Normality of a solution is defined as the number of gram equivalents of the solute dissolved per litre (dm3) of given solution. is V the volume of solution?
But the volume is already given as 10 ml, so it has to be something that's not the volume.
20 volume means 1dm3 of the H2O2 solution will produce 20dm3 of oxygen, where: H2O2 -> O2 + H2: i.e 1:1 stoichiometry. At 298K and 1atm pressure, 1 mole of gas occupies 24.45dm3. So 20/24.45 = 0.818 mol. O2 produced, which means there were 0.818mol H2O2 in 1dm3. Therefore molarity is 0.818M at 298K.
check this out @Abhisar i think this well help
will**
And it's an aqueous solution?
Wow! Thank you very much @Abmon98 !
your most welcome :D
Wow, nice find!
Thanx @ParthKohli !
\(\color{blue}{\text{Originally Posted by}}\) @ParthKohli And it's an aqueous solution? \(\color{blue}{\text{End of Quote}}\) Yes the solution is aqueous
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