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Chemistry 8 Online
OpenStudy (anonymous):

when 2.3g of C2H5OH is dissolved in 500 ml.The normality of solution will be??

OpenStudy (anonymous):

Mol.Wt of C2H50H=46 How to calculate normality??

OpenStudy (anonymous):

You know about equivalence weight ? @Ryaan

OpenStudy (anonymous):

\[\Large c=\frac{ n }{ v }\] c=concentration (Molarity) n=no. of moles v=volume

OpenStudy (anonymous):

To find n which is the number of moles, \[\Large n=\frac{ m }{ M }\] n=no. of moles m=mass M=Molecular weight

OpenStudy (anonymous):

Please explain me how to solve :/

OpenStudy (anonymous):

right Answer is 0.5..but when I divide m that is given mass by M molecular weight..answer comes 0.05..so tell me the complete procedure

OpenStudy (anonymous):

0.1 is correct answer * how it comes?

OpenStudy (anonymous):

to find the molarity, you need to find the number of moles because it wasn't given to you directly in the question so using the formula i stated above for finding the number of moles what do you get?

OpenStudy (anonymous):

aaaaaa maan its normality :/

OpenStudy (anonymous):

are you able to solve it or should i help you through with it?

OpenStudy (anonymous):

You help please

OpenStudy (anonymous):

first of all, let's find the number of moles. you were given the mass to be 2.3g and the molecular weight to be 46. plug it into the formula n=m/M...

OpenStudy (anonymous):

\[\Large n=\frac{ 2.3 }{ 46 }\] \[\Large n=??\]

OpenStudy (anonymous):

0.05

OpenStudy (anonymous):

you are right.... the question already gave you the volume to be 500ml.. all you have to do is to change that into liters.... are you able to do that?

OpenStudy (anonymous):

500x1/1000? is it right?

OpenStudy (anonymous):

=0.5

OpenStudy (anonymous):

sorry for the late reply...you are right! so you have 0.5L..

OpenStudy (anonymous):

now that we have the values needed we can put them in the formula for finding the molarity i stated above... \[\Large c=\frac{ 0.05~{\mathbb{mol}} }{ 0.5~{\mathbb{L}} }\] \[\Large c=...??...{\mathbb{M}}\]

OpenStudy (anonymous):

0.1 ..

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