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Chemistry 18 Online
OpenStudy (anonymous):

a solution of sucrose (c12h22o11) contains 13.5g in 100 ml of the solution the density =1.05kg/L calculate the molality of the solution

OpenStudy (anonymous):

@shivam_bhalla plzz help

OpenStudy (anonymous):

Show your working out/ attempt/ approach you have taken please.

OpenStudy (anonymous):

If you are totally clueless, then read yur textbook and come back again

OpenStudy (anonymous):

molality=no of moles/in 1kg of solvent

OpenStudy (anonymous):

Yes. So what is your no of moles ??

OpenStudy (anonymous):

moles=given wg/molar mass

OpenStudy (anonymous):

you know the molar mass of substance , right?? Calculate and tell me the molar mass

OpenStudy (anonymous):

mm=342

OpenStudy (anonymous):

right. Now how will you find given wt.?? Any ideas ??

OpenStudy (anonymous):

no

OpenStudy (anonymous):

(Tip : see the information given to you in the question that you haven't used yet)

OpenStudy (anonymous):

13.5

OpenStudy (anonymous):

and density. Right?? Now do you know the relation between density,mass and volume ??

OpenStudy (anonymous):

d=m/v

OpenStudy (anonymous):

Great. Now find your m from that relation and your m will be your given mass

OpenStudy (anonymous):

m=d*v

OpenStudy (anonymous):

So you will get your number of moles now. Next you know the volume of solution in litres and hence you know yourr molarity

OpenStudy (anonymous):

m=0.105

OpenStudy (anonymous):

Yes @ openstudy/ You found the given mass. Now follow what I told you in the last comment

OpenStudy (anonymous):

Sorry. There was a typo in my 2nd last comment

OpenStudy (anonymous):

ok then wat to do

OpenStudy (anonymous):

So now find your number of moles now.

OpenStudy (anonymous):

0.105/342= 3.05161 i am getting while i divided this.... i know this is not possible can u do it

OpenStudy (anonymous):

Ok. So now you found your number of moles(from above link). So what is your molality now??

OpenStudy (anonymous):

hey i got a number which is very small and it does not match to my option

OpenStudy (anonymous):

the answer should be ~0.43

OpenStudy (anonymous):

Let me check

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

@shivam_bhalla r an indian ....if u r which state??

OpenStudy (anonymous):

we have nt used 13.5 gm i think there is a mistake

OpenStudy (anonymous):

I got it now

OpenStudy (anonymous):

can u show that plz

OpenStudy (anonymous):

One question. The density is given for solution, right ??

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

density =1.05kg/L

OpenStudy (ujjwal):

Is the answer equal to 0.3947 ??

OpenStudy (anonymous):

Yes. We made a mistake in calculating given mass. The given mass is actually 13.5 gm

OpenStudy (anonymous):

the ans=0.43

OpenStudy (anonymous):

@ujjwal , you are making confusion in calculating wt of solvent

OpenStudy (anonymous):

the molality=0.39

OpenStudy (anonymous):

that is wrong.............

OpenStudy (anonymous):

Guys, wt of solvent is not 0.1 gm

OpenStudy (anonymous):

then wat???

OpenStudy (anonymous):

*kg

OpenStudy (ujjwal):

If ans is 0.39, use relation g/l =molarity*molecular wt You don't need density here g/l=13.5/0.1

OpenStudy (anonymous):

i think we have to do something with density

OpenStudy (ujjwal):

No.. we don't need density.. molarity has nothing to do with density.. use the relation i provided.. you will get answer 0.39 M

OpenStudy (anonymous):

it is molality

OpenStudy (ujjwal):

Volume is clearly stated to be 100ml=0.1l

OpenStudy (anonymous):

mass of solute = density * volume = 1.05 * 0.0135

OpenStudy (anonymous):

ok then

OpenStudy (anonymous):

I will have to sleep early. Sorry. @JFraser will help you

OpenStudy (anonymous):

I am not getting it at the moment . ;(

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