Question

The molarity and molality of a certain solution of acetic acid in water are 2.02 and 2.27 respectively Calculate the density?

Answer 1

@heena plzzz help

Answer 2

m= M/(d-MM2)

Answer 3

i think i have to use this am i right....

Answer 4

write the formula of
molarity thats is n/v
molality is mass of solute/mass of solvent
now u knw n=moles can be writeen in different wayby eeing u need density we will use densityxvolume/molecular weight

Answer 5

@open_study1
Show us your work and we will help.
Posting any formula such as: m= M/(d-MM2) is not enough.

Answer 6

2.27=2.03/(d-121.8)

Answer 7

is that right........

Answer 8

@open_study1 that is the ans u get or it is the formula?? sorry i m not familiar wid dat equation... so cant say anything can u show me how u get that ...

Answer 9

m= M/(d-MM2) m2 = molar mass of solute

Answer 10

ok and wat is M= is this molar mass of solvent? d=?? what is this

Answer 11

M=molarity m=molality

Answer 12

if u can plzz show ur way plzz

Answer 13

do u knw mass of solution??
well u kne by molarity u can find density as molarity
=n/v=density x v/v=density=2.02

Answer 14

@open_study1 plz pot the full qn dont try to make paradoz as in formula u put the value of m2 how u get that? there is no where given in qn :?

Answer 15

*paradox

Answer 16

ok plzz solve in ur way.....

Answer 17

well u didnt give me the whole info how can i solve say? u having m2 mean mas of solute as u said anything else u were provided??

Answer 18

let's assume that we have 1000ml of the solution, okay?

Molarity is 2.02. So that means there are 2.02 moles of acetic acid (CH3COOH) in it.

Now acetic acid's molar mass is 60 g/mol.
So, 1 liter of the solution contains 60*2.02 grams of acetic acid = 121.2 grams

Now, if the density of the solution is x g/ml, then it weighs 1000x grams.

Out of these 1000x grams, 121.2 grams is acetic acid. so the solvent is basically=(1000x-121.2)gms

Now molality is 2.27, and the no. of moles is present is 2.02.
Also, molality is calculated as the no. of moles of solute present per kg of the solution.


Can you work this out now?