Question

How do I calculate the molalities of 1.22M sugar (C12H22O11) solution density = 1.12g/mL (1.74m)

Answer 1

Hey @rgermina5873 and welcome to \(\LARGE \sf \bbox[#40B9E9]{\color{white}{Open}}\bbox[#A8CE91]{\color{white}{Study!}}\)

The molality \(b\) is defined as:\[\large b=\frac{ n }{ m }\]
We got a concentration and density defined as:\[\large C=\frac{ n }{ V }\] \[\large \rho = \frac{ m }{ V }\]
Lets see what happens when we multiply with the concentration with
the reciprocal density \(\rho^{-1}\):
\[\large C \times \rho^{-1}=\frac{ n }{ V } \times \frac{ V }{ m }=\frac{ n }{ m }=b\]
So we see that:
\[\large b=\frac{ C }{ \rho }\]
Make sure that the units get correct as molality has the unit \([\large \frac{mol}{kg}]\)

Answer 2

You are good. I could never understand this problem but you just get it. Thanks.

Answer 3

Thank you, but you understand how we derived our wished equation? :)

Answer 4

No not really. Please explain

Answer 5

Okay I try again.
molality \(b\) is a measure of how many moles of molecules we got per kilogram. The equation would therefor be:
\[\large b=\frac{ n }{ m }\]Where \(n\) is the amount of substance [mol] and m the mass (of the solvent) [kg]. The unit for molality therefor become \(\large \frac{mol}{g}\)

Understand so far? :)

Answer 6

Yes I understand so far.

Answer 7

Good we now look at the density and concentration:

The density \(\rho\) (the sign is called "rho") is defined as a measure how much a substance weights per volume. The equation become:
\[\large \rho = \frac{ m }{ V }\]
Where \(m\) is the mass [g] and \(V\) the volume [ml].

The concentration \(C\) is a measure of how many mole of molecules we got per volume. The equation become:
\[\large C=\frac{ n }{ V }\]
The \(n\) and \(V\) you know from before, but what is special this time is the unit: we define as new united called \(M\) and is equal to \(\frac{mol}{l}\)

Answer 8

So try multiply:
The reciprocal densitry \(\rho^{-1}\) and the concentration \(C\)
Note:
\[\large \rho ^{-1}=\frac{ 1 }{ \frac{ m }{ V } }=\frac{ V }{ m }\]

Answer 9

I've never seen that formula before.

Answer 10

Using this formula will help me get to the answer. Thanks once again.

Answer 11

Great work @Frostbite :)